A man is standing on top of a building 100 m high. He throws two balls vertically, one at $t = 0$ and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is +15 m at $t = 2$ s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.

A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other afte

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t = 0

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Concept understanding — Motion Under Gravity

Motion Under Gravity – The Intuition First

Drop a ball. It falls. Why? Because the Earth pulls it. But the key insight is not just that it falls — it's how it falls. If you drop a stone from a cliff, does it move at a constant speed? No. It gets faster as it goes down. That's the heart of motion under gravity: the pull is steady, so the acceleration is steady.

Now imagine throwing a ball straight up. It rises, slows down, stops for an instant at the top, then falls back down. The same pull that made it slow down on the way up makes it speed up on the way down. The acceleration never changes direction — it always points downward, toward the centre of the Earth.

That constant downward acceleration is called $g$ . Near the Earth's surface, its magnitude is about $9.8\ \text{m/s}^2$ (often taken as $10\ \text{m/s}^2$ in exams for quick calculation). The direction is always downward.

Important

In one-dimensional motion under gravity, the acceleration is constant and equal to $g$ downward. This is true only when air resistance is negligible and the height is small compared to the Earth's radius.

The Precise Statement

Motion under gravity (in one dimension) means an object moves vertically under the sole influence of the Earth's gravitational pull. The acceleration is:

a = -g

where the negative sign indicates downward direction (if we take upward as positive). The value of $g$ is $9.8\ \text{m/s}^2$ (or $10\ \text{m/s}^2$ in many problems).

Because acceleration is constant, all three equations of uniformly accelerated motion apply directly:

v = u + at

s = ut + \frac12 at^2

v^2 = u^2 + 2as

But here $a$ is replaced by $-g$ (if upward is positive) or $+g$ (if downward is positive). The choice of sign convention is yours — just be consistent.

v = u - gt

Key Features You Must Know

1. Free Fall (dropped from rest)

If you simply let go of an object ( $u = 0$ ), it falls with increasing speed. After time $t$ , its velocity is $v = gt$ (downward). The distance fallen is $s = \frac12 gt^2$ .

2. Projected Upward

If you throw a ball upward with speed $u$ , it rises until its velocity becomes zero at the highest point. That takes time $t = u/g$ . The maximum height reached is $h = u^2/(2g)$ .

3. Symmetry of the Motion

The time to go up equals the time to come back down (to the same height). The speed at which it returns to the launch point equals the initial speed $u$ (but downward). This symmetry is a direct consequence of constant acceleration.

Watch out

At the highest point, velocity is zero but acceleration is still $g$ downward. Many beginners think acceleration becomes zero at the top — it does not. The object is momentarily at rest, but gravity is still pulling it.

A Simple Example …

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Two balls thrown vertically from the same height maintain a constant separation after $t = 2\,\text{s}$ , which happens only when they have equal velocities. Using the 15 m gap at $t = 2\,\text{s}$ and the velocity relation $u_2 = u_1/2$ , we find $u_1 = 20\,\text{m/s}$ , $u_2 = 10\,\text{m/s}$ , and the time interval $\Delta t = 1\,\text{s}$ .

Why the gap becomes constant

When two objects move under gravity alone, their separation changes unless they share the same velocity. Think of it this way: if ball 1 is moving faster upward (or slower downward) than ball 2, the gap between them grows; if slower upward (or faster downward), the gap shrinks. A constant gap means both balls are moving at exactly the same speed at every instant after $t = 2\,\text{s}$ .

Since both balls experience the same acceleration $g = 10\,\text{m/s}^2$ downward, their velocities at time $t$ are:

v_1(t) = u_1 - gt, \quad v_2(t) = u_2 - g(t - \Delta t)

where $u_1$ and $u_2$ are the initial speeds, and ball 2 is thrown $\Delta t$ seconds after ball 1.

For the gap to remain constant from $t = 2\,\text{s}$ onward, we need $v_1(2) = v_2(2)$ :

u_1 - 10 \cdot 2 = u_2 - 10(2 - \Delta t)

u_1 - 20 = u_2 - 20 + 10\Delta t

u_1 = u_2 + 10\Delta t \quad \cdots (1)

We are also told $u_2 = u_1/2$ , so:

u_1 = \frac{u_1}{2} + 10\Delta t

\frac{u_1}{2} = 10\Delta t

u_1 = 20\Delta t \quad \cdots (2)

Finding the separation at $t = 2\,\text{s}$

The position of each ball at time $t$ (measured downward from the top of the building as positive) is:

s_1(t) = u_1 t - \frac{1}{2}g t^2

s_2(t) = u_2(t - \Delta t) - \frac{1}{2}g(t - \Delta t)^2 \quad \text{for } t \geq \Delta t

The vertical gap (ball 1 ahead, so $s_1 > s_2$ ) at $t = 2\,\text{s}$ is:

\Delta s = s_1(2) - s_2(2) = 15\,\text{m}

Substitute the positions:

u_1 \cdot 2 - 5 \cdot 4 - \left[u_2(2 - \Delta t) - 5(2 - \Delta t)^2\right] = 15

2u_1 - 20 - u_2(2 - \Delta t) + 5(2 - \Delta t)^2 = 15

2u_1 - u_2(2 - \Delta t) + 5(2 - \Delta t)^2 = 35 \quad \cdots (3)

Solving the system

From equation (2), $u_1 = 20\Delta t$ and $u_2 = u_1/2 = 10\Delta t$ .

Substitute into equation (3):

$2(20\Delta t) - 10\Delta t(2 - \Delta t) + 5(2 - \Delta t)^2 = 35$ …

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Concept: Constant Relative Velocity Under Equal Accelerations — Evaluate the Equal-Velocity Condition at the Natural Instant, Not at $t=2\,\text{s}$

Method: The Relative-Velocity-Invariant Shortcut — Both Conditions Read Off at the Moment Ball 2 is Thrown, Never Needing the Given $t=2\,\text{s}$ in the Algebra

The stored answer evaluates $v_1(2)=v_2(2)$ and separately builds full position expressions at $t=2\ \text{s}$ (with $(2-\tau)$ terms) to use the $15\ \text{m}$ gap. This method uses one structural fact — since both balls share the identical acceleration $-g$ once both are in flight, their relative velocity is constant for the entire time they're both airborne, not just from $t=2\ \text{s}$ onward — to derive both required equations at the single most natural moment: $t=\tau$ , the instant ball 2 is thrown, when ball 2's own position and velocity are both known immediately from its initial conditions.

Steps

State the key structural fact. For $t\ge\tau$ (both balls in flight), $v_1(t)=u_1-gt$ and $v_2(t)=u_2-g(t-\tau)$ , so their difference is:

v_1(t)-v_2(t) = \left[u_1-gt\right]-\left[u_2-g(t-\tau)\right] = u_1-u_2-g\tau

This expression has no $t$ in it at all — the relative velocity is automatically the same constant at every instant $t\ge\tau$ , simply because both balls share the same acceleration. "The gap remains constant" (i.e. $v_1-v_2=0$ ) is therefore either true for every $t\ge\tau$ at once, or false for all of them — there is nothing special about the given $t=2\ \text{s}$ in this condition.

So the equal-velocity condition can be evaluated at the SIMPLEST possible instant — $t=\tau$ itself — instead of at $t=2\ \text{s}$ . At $t=\tau$ : $v_1(\tau) = u_1-g\tau$ , and $v_2(\tau) = u_2$ (ball 2 has just been thrown, so $(t-\tau)=0$ and its velocity is exactly its initial speed). Setting them equal:

u_1 - g\tau = u_2 \quad\cdots(\star)

reached without ever substituting $t=2$ .

Use the given relation $u_2 = u_1/2$ directly in $(\star)$ :

u_1 - g\tau = \frac{u_1}{2} \;\Longrightarrow\; \frac{u_1}{2} = g\tau \;\Longrightarrow\; u_1 = 2g\tau = 20\tau \quad (g=10\ \text{m/s}^2)

Get the second equation the same way — evaluate the GAP itself at $t=\tau$ , not at $t=2\ \text{s}$ . At the instant ball 2 is thrown, its position is (by definition) exactly where it started: $y_2(\tau)=0$ . So the gap at that instant is simply

\text{gap}(\tau) = y_1(\tau) - y_2(\tau) = y_1(\tau) - 0 = y_1(\tau)

And because the gap is constant for all $t\ge\tau$ (established in Step 1, once $(\star)$ holds), this value at $t=\tau$ must equal the same constant $15\ \text{m}$ given at $t=2\ \text{s}$ — the problem's " $15\ \text{m}$ at $t=2\ \text{s}$ " and "the gap at the moment of the second throw" are one and the same number, precisely because the gap never changes once $(\star)$ is satisfied:

y_1(\tau) = u_1\tau - \frac{1}{2}g\tau^2 = 15

Solve the system. Substitute $u_1=20\tau$ from Step 3:

20\tau\cdot\tau - 5\tau^2 = 15 \;\Longrightarrow\; 20\tau^2-5\tau^2=15 \;\Longrightarrow\; 15\tau^2=15 \;\Longrightarrow\; \tau^2=1 \;\Longrightarrow\; \tau=1\ \text{s}

Then $u_1=20(1)=20\ \text{m/s}$ , and $u_2=u_1/2=10\ \text{m/s}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. If an object is thrown vertically upward, then at the highest point its velocity will be$
(a) maximum
(b) same as initial velocity
(c) zero
(d) none of these
›Reveal solutionSolution

At the highest point of upward motion the velocity is momentarily zero. Answer (C).

When an object is thrown vertically upward, gravity decelerates it at g. Its speed keeps decreasing until, at the highest point, it becomes zero (v = u - gt reaches 0 when t = u/g).

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which one is uniform of a body falling freely under the action of gravity? (A) Momentum (B) Acceleration (C) Velocity (D) Speed$
›Reveal solutionSolution
Acceleration is the uniform (constant) quantity during free fall; velocity, speed, and momentum all keep changing.

For a body falling freely under gravity (ignoring air resistance):

Acceleration $a = g$ , a fixed constant value at a given place — does not change with time.

Velocity $v = u + gt$ — increases continuously with time. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. A stone dropped from the bridge reaches the water in 4 seconds. The height of the bridge is$
(a) 78.4 m
(b) 2 m
(c) 260 m
(d) 32 m
›Reveal solutionSolution

Free fall from rest: h = (1/2) g t^2 = (1/2)(9.8)(4)^2 = 78.4 m.

The stone is dropped (initial velocity u = 0) and falls freely under gravity for t = 4 s.

Using the second equation of motion: h = u*t + (1/2) g t^2

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A body dropped from the top of the tower of height H meters. It takes 'T' time to reach the ground. Where is the body T/2 time after the release?$
(a) At H/2 meters from the ground
(b) At H/4 meters from the ground
(c) At 3H/4 meters from the ground
(d) Depends on mass and volume of body
›Reveal solutionSolution

At half the total fall time, the body has covered only a quarter of the total height (not half), because distance under free fall grows with $t^2$ , not $t$ .

Let the body fall from height $H$ and reach the ground in time $T$ . Using $H = (1/2)g T^2$ ... (1)

Distance fallen in time $T/2$ : $h = (1/2)g(T/2)^2 = (1/2)g \cdot T^2/4 = (1/4)\left[(1/2)gT^2\right] = H/4$

So after $T/2$ seconds, the body has fallen $H/4$ metres. Its height above the ground at that instant is:

$H - H/4 = 3H/4$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. An object is dropped in a planet from height 50 m, it reaches the ground in 2 s. The acceleration due to gravity in the planet is:$
(a) g = 15 ms^-2
(b) g = 20 ms^-2
(c) g = 30 ms^-2
(d) g = 25 ms^-2
›Reveal solutionSolution

Using h = (1/2) g t^2 for an object dropped from rest, h = 50 m and t = 2 s give g = 25 ms^-2.

When an object is 'dropped' (released from rest, initial velocity u = 0) and falls under a constant gravitational acceleration g, the distance fallen in time t is given by the standard kinematic equation:

h = u t + (1/2) g t^2 = (1/2) g t^2 (since u = 0)

Substitute the given values h = 50 m, t = 2 s:

50 = (1/2) x g x (2)^2 …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. A body falling freely under gravity has uniform (A) Speed (B) Velocity (C) Momentum (D) Acceleration$
›Reveal solutionSolution

Free fall under gravity has uniform acceleration $g$ , not uniform speed/velocity/momentum.

During free fall, the only force is gravity (weight $mg$ , ignoring air resistance), which is constant. By Newton's second law, acceleration $=F/m=g$ , a constant value throughout the fall. Speed, veloci …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. A ball is projected vertically upwards with a velocity v. It comes back to ground in time t. Which of the following v-t graph shows the motion correctly?$
(a) graph
(a)
(b) graph
(b)
(c) graph
(c)
(d) graph
(d) (The four option graphs are shown in the original question paper.)
›Reveal solutionSolution

Under constant gravitational deceleration, velocity changes linearly with time throughout the whole up-and-down motion, so the v-t graph must be a single straight line, not a V-shape or triangle shape.

When a ball is projected vertically upward with initial velocity v and returns to the same point (the ground) in time t, its acceleration is constant throughout the motion: a = -g (taking upward as positive), because gravity acts continuously, both while going up and coming down — there is no sudden change in acceleration at the highest point.

Since a = dv/dt is constant, v must vary linearly with time: v(t) = v - g t.

At t = 0, v = +v (initial upward velocity, positive intercept).

At t = t/2 (roughly midway), the ball reaches its highest point, so v = 0 — the line crosses the t-axis there. …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. A football is kicked into the air vertically upwards with velocity u. The velocity of the ball at the highest point is$
(1) u
(2) 2u
(3) zero
(4) 4u
›Reveal solutionSolution

At the highest point of vertical projectile motion, the velocity is instantaneously zero, since that is precisely the point where the upward motion stops and downward motion begins.

When the football is kicked straight up with velocity u, gravity decelerates it at g = 9.8 m/s^2 (downward). Using v = u - gt, the velocity decreases steadily until it becomes zero - this instant defines the 'highest point', because after this the velocity becomes negative (i.e. the ball starts moving downward). So by d …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. If an object is falling from a height of 20 m, then the time taken by the object to reach the ground : (ignore air resistance and take g = 10 ms^-2)$
(a) 2 s
(b) 1.732 s
(c) 1.532 s
(d) 1.414 s
›Reveal solutionSolution

Using h = (1/2) g t^2 for a body falling from rest, t = sqrt(2h/g) = 2 s.

Given: h = 20 m, g = 10 m/s^2, initial velocity u = 0 (the object is falling, and we ignore air resistance).

Using the second equation of motion, …

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CBSE 2022Set TERM11 markMCQ
$Q. Free fall of an object in vacuum is a case of motion with$
(1) uniform velocity
(2) uniform acceleration
(3) variable acceleration
(4) uniform speed
›Reveal solutionSolution

With no air resistance to vary the net force, the only force on a freely falling object is gravity, which produces a constant acceleration g -- hence uniform (not variable) acceleration.

In vacuum, the only force acting on the falling object is its weight, mg, which is constant near the Earth's surface. By Newton's second law, acceleration = F/m = g, a constant value (~9.8 m/s^2) independent of the object's mass. Since this acceleration does not change with time, the motion is one …

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CBSE 2021Set TERM11 markMCQ
$Q. A ball thrown by one player reaches the other in 2 sec. The maximum height attained by the ball above the point of projection will be (take g=10 ms^-2).$
(a) 10 m
(b) 7.5 m
(c) 5 m
(d) 5.5 m
›Reveal solutionSolution

Total time of flight = 2 s, so time to reach the highest point = 1 s. At the highest point vertical velocity = 0, which gives initial speed u = g x 1 = 10 m/s, and then h = ut - (1/2)gt^2 = 5 m.

Since the ball leaves one player and reaches the other at (essentially) the same height, its up-and-down motion is symmetric about the highest point of the path. So:

time to rise to maximum height, t = T/2 = 2/2 = 1 s

…

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CBSE 2020Set ANNUAL1 markMCQ
$Q. A ball is dropped from the top of a tower and reaches the ground in 4 seconds. The height of the tower is: (g = 10 m/s^2)$
(a) 40 m
(b) 60 m
(c) 80 m
(d) 100 m
›Reveal solutionSolution

A dropped ball starts from rest, so its fall height follows h = (1/2) g t^2; plugging in t = 4 s gives 80 m.

…

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Why the gap becomes constant

Finding the separation at t=2 st = 2\,\text{s}t=2s

Solving the system

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Finding the separation at $t = 2\,\text{s}$

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