Q.A stone dropped from the top of a tower of height $300\ \text{m}$ splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is $340\ \text{m s}^{-1}$? ($g = 9.8\ \text{m s}^{-2}$)

Showing the 12 most recent of 14 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.A body is falling freely under gravity from a height of 200 m. The total displacement of the body during the second half-second, fourth half-second and sixth half-second of its motion is (acceleration due to gravity $= 10\ ms^{-2}$) (A) 26.25 m (B) 32.50 m (C) 37.25 m (D) 42.25 m

   ›Reveal solutionSolution

   Summing the displacement in the 2nd, 4th and 6th half-second intervals of free fall (using $s=\frac12g(t_2^2-t_1^2)$ for each) gives a total of $26.25$ m.

   Concept and Intuition

   For motion starting from rest under constant acceleration, the displacement in any time window $[t_1,t_2]$ is simply $s = \frac12 g(t_2^2-t_1^2)$ — a direct consequence of integrating $v=gt$. Rather than computing position at every instant, we can directly evaluate this formula for each of the three required half-second windows and add them.

   Step-by-Step Solution

   1. Identify the half-second windows (each is $0.5$ s wide), counting from $t=0$:
      • 1st: $[0, 0.5]$s, 2nd: $[0.5, 1.0]$s, 3rd: $[1.0,1.5]$s, 4th: $[1.5,2.0]$s, 5th: $[2.0,2.5]$s, 6th: $[2.5,3.0]$s.
   2. Use $s = \dfrac12 g(t_2^2-t_1^2)$ with $g=10\ ms^{-2}$, so $\dfrac12 g = 5$.
   3. 2nd half-second, $t_1=0.5,\ t_2=1.0$: $s_2 = 5(1.0^2-0.5^2)=5(1-0.25)=5(0.75)=3.75$ m.
   4. 4th half-second, $t_1=1.5,\ t_2=2.0$: $s_4 = 5(2.0^2-1.5^2)=5(4-2.25)=5(1.75)=8.75$ m.
   5. 6th half-second, $t_1=2.5,\ t_2=3.0$: $s_6 = 5(3.0^2-2.5^2)=5(9-6.25)=5(2.75)=13.75$ m.
   6. Total displacement $= 3.75+8.75+13.75 = 26.25$ m.
   7. Sanity check on validity: the body falls from a height of 200 m; the time to reach the ground is $t=\sqrt{2\times200/10}=\sqrt{40}\approx6.32$ s, so at $t=3$ s (end of the 6th half-second) the body has fallen only $\frac12(10)(3)^2=45$ m — well short of 200 m, confirming it is still in free fall throughout and the calculation is valid.

   Common Mistakes

   • Using average velocity times time incorrectly instead of the exact $\frac12g(t_2^2-t_1^2)$ formula, which can introduce small arithmetic errors.
   • Miscounting which half-second interval is "the 4th" — it's easy to off-by-one the interval boundaries (e.g., confusing $[1.0,1.5]$ with $[1.5,2.0]$).

   ✓Final answer

   The correct option is (A) — 26.25 m.

   ANSWER: A

2. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If the distance travelled by a freely falling body in the last but one second of its motion is 5 m, then the last second is (Acceleration due to gravity $= 10 \mathrm{m\,s^{-2}}$) (A) $1^{st}$ (B) $2^{nd}$ (C) $3^{rd}$ (D) $4^{th}$

   ›Reveal solutionSolution

   This tests the "distance in the $n$th second" formula for free fall from rest, then back-solving for the total fall duration. Answer: (B).

   Concept and Intuition

   For a body released from rest, the distance covered in each successive second grows linearly (an arithmetic progression) because speed keeps increasing uniformly. The formula $s_n = u + \frac{g}{2}(2n-1)$ (here $u=0$) captures this directly, letting us go from "distance in a given second" back to "which second it was."

   Step-by-Step Solution

   1. Distance fallen in the $n$th second, starting from rest: $s_n = \dfrac{g}{2}(2n-1)$. With $g=10\,\text{m s}^{-2}$, $s_n = 5(2n-1)$.
   2. Let the total time of fall be $T$ seconds (so the body lands at $t=T$). The last second of motion is the interval $n=T$; the last-but-one second is $n=T-1$.
   3. We're told $s_{T-1} = 5\,\text{m}$: $5\big(2(T-1)-1\big) = 5 \implies 2(T-1)-1 = 1 \implies 2(T-1)=2 \implies T-1=1 \implies T=2$.
   4. So $T=2$ s, meaning the last-but-one second is the 1st second (consistent: $s_1 = 5(1)=5\,$m ✓), and the last second is the 2nd second.

   Common Mistakes

   • Mixing up "last second" with "last but one second" — the last-but-one is one interval before the final one, i.e. $n=T-1$, not $n=T$.
   • Assuming $u\neq0$; the problem is "freely falling," so it starts from rest.

   ✓Final answer

   The correct option is (B) — $2^{nd}$.

   ANSWER: B

3. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.If a ball released from a height H takes a time T to reach the ground, then the position of the ball from the ground at a time $\dfrac{T}{2}$ is (A) $\dfrac{H}{4}$ (B) $\dfrac{H}{2}$ (C) $\dfrac{3H}{4}$ (D) $\dfrac{2H}{3}$

   ›Reveal solutionSolution

   At the halfway point in time, the ball has fallen only a quarter of the total height (since distance fallen from rest scales as $t^2$) — so it's still at $3H/4$ above the ground. Answer: (C).

   Concept and Intuition

   For free fall starting from rest, distance fallen grows with the SQUARE of elapsed time, not linearly. This means the ball covers a much smaller fraction of its total fall distance in the first half of its total fall time than in the second half — a classic conceptual trap in kinematics.

   Step-by-Step Solution

   1. Total fall relation: $H = \tfrac12 g T^2$ (since initial velocity is zero, dropped from rest).
   2. Distance fallen at time $t = T/2$: $d = \tfrac12 g \left(\dfrac{T}{2}\right)^2 = \tfrac12 g \cdot \dfrac{T^2}{4} = \dfrac{1}{4}\left(\tfrac12 g T^2\right) = \dfrac{H}{4}$.
   3. So at $t = T/2$, the ball has fallen $H/4$ from its starting point.
   4. Its remaining height above the ground is therefore $H - \dfrac{H}{4} = \dfrac{3H}{4}$.
   5. Hence, the position of the ball from the ground at time $T/2$ is $3H/4$.

   Common Mistakes

   • Assuming that at half the total time, the ball has fallen half the total height (that would be true only for constant-velocity motion, not accelerated free fall from rest).
   • Confusing 'distance fallen' with 'distance remaining above the ground' — the question asks for the position (height above ground), which is $H$ minus the distance fallen, not the distance fallen itself.

   ✓Final answer

   The correct option is (C) — $\dfrac{3H}{4}$.

   ANSWER: C

4. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.A body is dropped freely from a height of 8 m from the ground. If the coefficient of restitution between the body and the ground is 0.5, then the maximum height reached by the body after second impact with the ground is (A) 2 m (B) 0.25 m (C) 1 m (D) 0.5 m

   ›Reveal solutionSolution

   Each bounce height is reduced by a factor $e^2$ from the previous drop height; two successive bounces from 8 m with $e=0.5$ give $0.5$ m after the second impact.

   Concept and Intuition

   The coefficient of restitution $e$ relates the speed of separation to the speed of approach at a collision: $e = \dfrac{\text{speed after}}{\text{speed before}}$. Since speed just before impact from height $h$ is $\sqrt{2gh}$, and the rebound speed is $e\sqrt{2gh}$, the rebound height is $h' = \dfrac{(e\sqrt{2gh})^2}{2g} = e^2 h$. This factor $e^2$ applies afresh at every successive bounce, using the height attained just before that particular bounce.

   Step-by-Step Solution

   1. Initial drop height: $h_0 = 8\ \mathrm{m}$.
   2. After the first impact, rebound height: $h_1 = e^2 h_0 = (0.5)^2 \times 8 = 0.25\times 8 = 2\ \mathrm{m}$.
   3. The body then falls back down from $h_1=2\,\mathrm{m}$ and undergoes a second impact; the rebound height after this second impact is: $h_2 = e^2 h_1 = 0.25\times 2 = 0.5\ \mathrm{m}$.
   4. So the maximum height reached after the second impact is $0.5\ \mathrm{m}$.

   Common Mistakes

   • Using $e$ (not $e^2$) directly on the height instead of on the velocity (then squaring for height).
   • Computing only the height after the first bounce instead of continuing to the second.

   ✓Final answer

   The correct option is (D) — 0.5 m.

   ANSWER: D

5. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If a stone thrown vertically upwards from a bridge with an initial velocity of $5\ \mathrm{m\,s^{-1}}$, strikes the water below the bridge in a time of 3 s, then the height of the bridge above the water surface is (Acceleration due to gravity $= 10\ \mathrm{m\,s^{-2}}$) (A) 10 m (B) 26 m (C) 30 m (D) 18 m

   ›Reveal solutionSolution

   This tests using the sign convention in kinematics equations for a projectile thrown upward that ends up below its starting point. The bridge height above the water is 30 m, option (C).

   Concept and Intuition

   When a stone is thrown upward but eventually lands below the launch point (like off a bridge into water below), we can still use $s=ut-\tfrac12gt^2$ with a single consistent sign convention (say, up = positive) for the entire motion — the equation automatically accounts for the stone going up, coming back down past the launch point, and continuing below it. The final displacement will simply come out negative, telling us the stone ended up below where it started.

   Step-by-Step Solution

   1. Take upward as positive. Initial velocity $u=+5\text{ m/s}$, acceleration $a=-g=-10\text{ m/s}^2$ (gravity acts downward), time $t=3\text{ s}$.
   2. Displacement: $s=ut+\tfrac12at^2=5(3)+\tfrac12(-10)(3)^2=15-45=-30\text{ m}$.
   3. The displacement is $-30$ m, meaning the stone's final position is 30 m below the bridge (the launch point).
   4. Since the stone "strikes the water below the bridge" at this final position, the water surface is 30 m below the bridge — i.e., the height of the bridge above the water is $30$ m.

   Common Mistakes

   • Splitting the motion into "going up" and "coming down" phases separately and mismanaging the bridge-to-water distance versus the total path length traveled (up + down) — the single-equation approach with signed displacement avoids this entirely.
   • Sign errors: using $g=+10$ throughout without keeping direction consistent, which can flip the final sign and produce the wrong magnitude interpretation.

   ✓Final answer

   The correct option is (C) — 30 m.

   ANSWER: C

6. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.A body falling freely under gravity from rest from certain height reaches the ground in a time 5 s. The distance travelled by the body in the last two seconds of its motion is (A) 98 m (B) 44.1 m (C) 58.8 m (D) 78.4 m

   ›Reveal solutionSolution

   This tests the equation $h=\tfrac12 g t^2$ for free fall and the idea of finding a distance interval by subtracting two cumulative distances.

   Concept and Intuition

   A body dropped from rest speeds up continuously, so it covers more distance in later seconds than in earlier ones. To find the distance covered in a specific time interval (here, the last 2 s of a 5 s fall), the cleanest method is to compute the total distance fallen at each of the two boundary times and subtract — rather than trying to build a "distance in nth second" formula for a 2-second window.

   Step-by-Step Solution

   1. Total fall time is $5$ s, so at $t=5$ s the total distance is $h(5)=\tfrac12 g t^2=\tfrac12(9.8)(5)^2=\tfrac12(9.8)(25)=122.5\ \text{m}$.
   2. The body reaches the ground at $t=5$ s, so "the last two seconds" means the interval from $t=3$ s to $t=5$ s. At $t=3$ s, $h(3)=\tfrac12(9.8)(3)^2=\tfrac12(9.8)(9)=44.1\ \text{m}$.
   3. Distance travelled in the last 2 s $=h(5)-h(3)=122.5-44.1=78.4\ \text{m}$.

   Common Mistakes

   • Using the "distance in $n$th second" formula $u+\tfrac{g}{2}(2n-1)$ for $n=5$ alone, which gives the distance in the 5th second (1 s window), not the last 2 seconds.
   • Forgetting $u=0$ since the body starts from rest.

   ✓Final answer

   The correct option is (D) — 78.4 m.

   ANSWER: D

7. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.A ball falls freely from rest on to a hard horizontal floor and repeatedly bounces. If the velocity of the ball just before the first bounce is $7\ \text{ms}^{-1}$ and the coefficient of restitution is 0.75, the total distance travelled by the ball before it comes to rest is (acceleration due to gravity = $10\ \text{ms}^{-2}$) (A) 10.75 m (B) 9.75 m (C) 8.75 m (D) 11.75 m

   ›Reveal solutionSolution

   Using the standard bouncing-ball series with coefficient of restitution $e=0.75$, the total distance travelled before the ball comes to rest is $8.75$ m.

   Concept and Intuition

   Each bounce reduces the rebound speed by a factor $e$ (coefficient of restitution), so the height reached after the $n$-th bounce is $h_0e^{2n}$ (since height $\propto$ speed$^2$). The ball travels down $h_0$ once, then up-and-down $2h_n$ for every subsequent bounce height $h_n$, forming a geometric series that sums to a finite total distance even though there are infinitely many bounces.

   Step-by-Step Solution

   1. Speed just before first bounce $v=7\ \text{ms}^{-1}$, so the initial fall height $h_0=\dfrac{v^2}{2g}=\dfrac{49}{20}=2.45$ m.
   2. After each bounce, rebound speed $=e\times$(impact speed), so rebound height $=e^2\times$(previous height).
   3. Heights after successive bounces: $h_0e^2, h_0e^4, h_0e^6,\dots$
   4. Total distance $=h_0+2(h_0e^2+h_0e^4+\cdots)=h_0+\dfrac{2h_0e^2}{1-e^2}=h_0\cdot\dfrac{1+e^2}{1-e^2}$.
   5. With $e=0.75=3/4$: $e^2=9/16$, so $\dfrac{1+9/16}{1-9/16}=\dfrac{25/16}{7/16}=\dfrac{25}{7}$.
   6. Total distance $=2.45\times\dfrac{25}{7}=\dfrac{61.25}{7}=8.75$ m.

   Common Mistakes

   • Forgetting the factor of $2$ for each bounce (the ball goes up and comes back down after every bounce except the very first descent).
   • Using $e$ (not $e^2$) directly as the height-reduction factor — height scales as the square of speed.

   ✓Final answer

   The correct option is (C) — 8.75 m.

   ANSWER: C

8. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.A body thrown vertically upwards from the ground reaches a maximum height 'H'. The ratio of the velocities of the body at heights $\frac{3H}{4}$ and $\frac{8H}{9}$ from the ground is (A) 4:9 (B) 27:32 (C) 3:2 (D) 3:8

   ›Reveal solutionSolution

   Use energy conservation (or kinematics) to express speed as a function of height below the maximum, then take the ratio at the two given heights. Answer: 3:2.

   Concept and Intuition

   For a body under gravity with maximum height $H$, its speed at any height $h$ (on the way up or down) is given by $v=\sqrt{2g(H-h)}$ — this follows directly from $v^2 = u^2-2gh$ combined with $u^2=2gH$ (from the condition that velocity is zero at height $H$).

   Step-by-Step Solution

   1. Speed at height $h$: $v(h)=\sqrt{2g(H-h)}$.
   2. At $h_1=\dfrac{3H}{4}$: $H-h_1=\dfrac{H}{4}$, so $v_1=\sqrt{2g\cdot H/4}=\sqrt{\dfrac{gH}{2}}$.
   3. At $h_2=\dfrac{8H}{9}$: $H-h_2=\dfrac{H}{9}$, so $v_2=\sqrt{2g\cdot H/9}=\sqrt{\dfrac{2gH}{9}}$.
   4. Ratio: $\dfrac{v_1}{v_2}=\sqrt{\dfrac{gH/2}{2gH/9}}=\sqrt{\dfrac{9}{4}}=\dfrac32$.
   5. So $v_1:v_2 = 3:2$.

   Common Mistakes

   • Forgetting that the relevant quantity for speed is $(H-h)$, the remaining height to rise, not $h$ itself.
   • Arithmetic slip simplifying the ratio of fractions under the square root.

   ✓Final answer

   The correct option is (C) — 3:2.

   ANSWER: C

9. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.A diving board is at a height of 'h' from the water surface. A swimmer standing on this board thrown a stone vertically upward with a velocity $16\ \mathrm{ms^{-1}}$. It reaches the water surface in a time of 5 s. In the next 0.2s, the diver can hear the sound from water surface. The speed of sound is (acceleration due to gravity $g = 10\ \mathrm{ms^{-2}}$) (A) $450\ \mathrm{ms^{-1}}$ (B) $225\ \mathrm{ms^{-1}}$ (C) $200\ \mathrm{ms^{-1}}$ (D) $275\ \mathrm{ms^{-1}}$

   ›Reveal solutionSolution

   Use kinematics to get the board's height above water from the stone's 5 s flight, then divide that height by the 0.2 s sound-travel time to get the speed of sound.

   Concept and Intuition

   This is a two-stage problem: first a projectile-motion calculation to pin down the (otherwise unknown) height $h$ of the diving board, then a simple distance/time calculation for sound, since the sound of the splash has to travel back up through air over the same vertical distance $h$ that the stone fell.

   Step-by-Step Solution

   1. Take upward as positive. The stone starts at the board (height $h$ above water) with $u=+16\,\text{ms}^{-1}$, acceleration $-g=-10\,\text{ms}^{-2}$, and reaches the water ($displacement=-h$ relative to the board) at $t=5\,\text{s}$.
   2. Using $s=ut+\tfrac12at^2$: $-h=16(5)-\tfrac12(10)(5)^2=80-125=-45$.
   3. So $h=45\,\text{m}$ — the board is 45 m above the water.
   4. The splash sound, generated at the water surface, must travel this same 45 m vertically back up to the diver's ears, taking the stated $0.2\,\text{s}$.
   5. Speed of sound $=\dfrac{\text{distance}}{\text{time}}=\dfrac{45}{0.2}=225\,\text{ms}^{-1}$.

   Common Mistakes

   • Forgetting that the stone is thrown UPWARD first (not just dropped), which is essential to correctly set up the kinematics equation with $u=+16$ and a negative net displacement.
   • Assuming the total 5.2 s (5 s + 0.2 s) is the sound's travel time instead of just the extra 0.2 s.

   ✓Final answer

   The correct option is (B) — $225\,\mathrm{ms^{-1}}$.

   ANSWER: B

10. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
    Q.From an elevated point 'A', a stone is projected vertically upwards. The velocity of the stone when it reaches a distance 'h' below 'A' is double its velocity when it was at a height 'h' above 'A'. The greatest height attained by the stone above 'A' is (A) $\dfrac{6h}{5}$ (B) $\dfrac{5h}{3}$ (C) $2h$ (D) $\dfrac{7h}{2}$

    ›Reveal solutionSolution

    Applying $v^2 = u^2 - 2gy$ at $y=+h$ and $y=-h$, and using the given speed ratio, the maximum height above A works out to $\dfrac{5h}{3}$. Answer: (B).

    Concept and Intuition

    For vertical motion under gravity, speed at any displacement $y$ from the launch point (taking up as positive) follows $v^2 = u^2 - 2gy$ regardless of the direction of travel — the same speed magnitude occurs at height $h$ on the way up and on the way down, but a greater speed occurs at a point below the launch point (since the object has fallen further under gravity from its highest point).

    Step-by-Step Solution

    1. Let $u$ be the launch speed at A (taking upward direction as positive, A as origin).
    2. Speed at height $+h$ above A: $v_1^2 = u^2 - 2gh$.
    3. Speed at $-h$ (i.e., $h$ below A, reached after the stone falls back past A): $v_2^2 = u^2 - 2g(-h) = u^2 + 2gh$.
    4. Given condition: $v_2 = 2v_1 \Rightarrow v_2^2 = 4v_1^2$.
    5. Substitute: $u^2 + 2gh = 4(u^2 - 2gh) = 4u^2 - 8gh$.
    6. Rearranging: $10gh = 3u^2 \Rightarrow u^2 = \dfrac{10gh}{3}$.
    7. Maximum height above A: $H = \dfrac{u^2}{2g} = \dfrac{10gh}{3 \times 2g} = \dfrac{5h}{3}$.

    Common Mistakes

    • Forgetting that the speed below the launch point is larger than at an equal height above it (because the stone has effectively "fallen" from its peak all the way to $-h$).
    • Sign errors when substituting $y = -h$ into $v^2 = u^2 - 2gy$.

    ✓Final answer

    The correct option is (B) — $\dfrac{5h}{3}$.

    ANSWER: B

11. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.The time of flight of a vertically projected stone is 8 s. The position of the stone after 6 s from the ground is (Acceleration due to gravity $= 10\ \mathrm{ms}^{-2}$) (A) 20 m (B) 60 m (C) 75 m (D) 40 m

    ›Reveal solutionSolution

    With total flight time 8 s, the initial speed is 40 m/s; plugging $t=6$ s into the standard kinematics equation gives a height of 60 m above the ground.

    Concept and Intuition

    For a vertically projected object under gravity, the total time of flight is symmetric: time to go up equals time to come down. Knowing the total flight time lets us find the initial speed, after which we can find the position (height) at any given time using the standard equation of motion.

    Step-by-Step Solution

    1. Since the stone is thrown vertically up and returns to the same level, time of flight $T = \dfrac{2u}{g}$, where $u$ is the initial speed.

    $8 = \frac{2u}{10} \implies u = 40\ \text{m/s}$

    2. Use the position equation (taking upward as positive, starting from the ground):

    $h(t) = ut - \frac12 g t^2$

    3. Substitute $t=6$ s:

    $h(6) = 40(6) - \frac12(10)(6)^2 = 240 - 5(36) = 240-180 = 60\ \text{m}$

    4. Sanity check via the peak: time to peak is $t_{peak}=u/g=4$ s, with peak height $=u^2/(2g)=1600/20=80$ m. At $t=6$ s (2 seconds past the peak, falling), height $=80-\frac12(10)(2)^2=80-20=60$ m — matches.

    Common Mistakes

    • Using $t=6$ s directly in $h=\frac12gt^2$ (a free-fall-from-rest formula), forgetting the stone was thrown upward with initial speed $u\ne0$.
    • Mixing up total flight time with time-to-peak (they differ by a factor of 2).

    ✓Final answer

    The correct option is (B) — 60 m.

    ANSWER: B

12. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.Two bodies of masses '$m_1$' and '$m_2$' are dropped from two different heights $h_1$ and $h_2$ respectively. The ratio of the times taken by the two masses to touch the ground is (neglect air resistance) (A) $\dfrac{h_1}{h_2}$ (B) $\dfrac{m_1 h_1}{m_2 h_2}$ (C) $\dfrac{m_1 h_2}{m_2 h_1}$ (D) $\sqrt{\dfrac{h_1}{h_2}}$

    ›Reveal solutionSolution

    Free-fall time depends only on the drop height (not mass), so the ratio of times is simply $\sqrt{h_1/h_2}$.

    Concept and Intuition

    A classic result of Galileo's: all objects, regardless of mass, fall with the same acceleration $g$ in the absence of air resistance. So the time to fall a height $h$ depends only on $h$ and $g$, never on the mass.

    Step-by-Step Solution

    1. For a body dropped (initial velocity 0) from height $h$: $h = \dfrac{1}{2}gt^2$.
    2. Solve for $t$: $t = \sqrt{\dfrac{2h}{g}}$.
    3. For the two bodies: $t_1 = \sqrt{2h_1/g}$, $t_2=\sqrt{2h_2/g}$.
    4. Ratio: $\dfrac{t_1}{t_2} = \sqrt{\dfrac{h_1}{h_2}}$ — the mass cancels out entirely and doesn't even appear.

    Common Mistakes

    • Trying to bring $m_1,m_2$ into the ratio (a very common trap in the options) — mass plays no role in free-fall time.

    ✓Final answer

    The correct option is (D) — $\sqrt{\dfrac{h_1}{h_2}}$.

    ANSWER: D