Q.A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800m from the foot of hill and can be moved on the ground at a speed of 2 m/s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill ? Take $g = 10$ m/s$^2$.

The maximum distance for satisfactory communication in line-of-sight (LOS) mode is limited by the curvature of the Earth. Radio waves in LOS mode travel in straight lines, so they cannot bend around the Earth's curve. An antenna can only transmit or receive signals up to its visible horizon.

The total maximum distance between a transmitting antenna and a receiving antenna is the sum of the maximum distances each antenna can "see" to the horizon.

Consider an antenna of height $h$ above the Earth's surface. Let $R$ be the radius of the Earth. The maximum distance $d$ from the antenna to the horizon can be found using the Pythagorean theorem. Imagine a right-angled triangle formed by:

1. The center of the Earth.
2. The point on the Earth's surface directly below the antenna.
3. The point on the horizon where the line of sight from the antenna touches the Earth tangentially.

The sides of this triangle are:

• One leg is the radius of the Earth, $R$, extending from the center to the horizon point.
• The other leg is the distance $d$ from the antenna to the horizon. This line is tangent to the Earth's surface at the horizon point, making it perpendicular to the radius $R$.
• The hypotenuse is the distance from the center of the Earth to the top of the antenna, which is $R+h$.

Applying the Pythagorean theorem:

$d^2 + R^2 = (R+h)^2$

$d^2 + R^2 = R^2 + 2Rh + h^2$

$d^2 = 2Rh + h^2$

Since the antenna height $h$ is typically much smaller than the Earth's radius $R$ ($h \ll R$), the term $h^2$ is negligible compared to $2Rh$.

Therefore, we can approximate:

$d^2 \approx 2Rh$

$d \approx \sqrt{2Rh}$

If we have a transmitting antenna of height $h_T$ and a receiving antenna of height $h_R$, the maximum distance for communication between them, $d_{max}$, is the sum of their individual horizon distances:

$d_{max} = d_T + d_R = \sqrt{2Rh_T} + \sqrt{2Rh_R}$

Now, let's apply this to the given problem.

1. Identify the given values and ensure consistent units.
   • Height of transmitting antenna, $h_T = 320 \text{ m}$
   • Height of receiving antenna, $h_R = 20 \text{ m}$
   • Radius of the Earth, $R = 6400 \text{ km} = 6400 \times 10^3 \text{ m}$ …

The key concept here is limiting friction. When a body is just about to move, the applied horizontal force equals the maximum static friction (limiting friction). The limiting friction $f$ is related to the normal reaction $N$ by $f = \mu N$, where $\mu$ is the coefficient of static friction. The weight of the body acts vertically downward, and on a horizontal surface, the normal reaction equals the weight unless there is an additional vertical component from the applied force. Here, the applied force is horizontal (as implied by "minimum force to move" on a horizontal surface), so the normal reaction is simply the weight minus any vertical pull — but since no angle is given, we assume the force is horizontal. That means $N = \text{weight} = 50 \, \text{N}$ initially seems plausible, but then the friction would be 28.28 N, which doesn’t match any option. So there must be a vertical component: the applied force is likely at an angle, reducing the normal reaction. The minimum force to move a body on a horizontal surface is achieved when the force is applied at an angle that optimizes the trade-off between reducing normal reaction and maintaining a useful horizontal component.

Let’s work through it step by step.

1. Set up the forces.

   Let the applied force $F = 28.28 \, \text{N}$ be at an angle $\theta$ above the horizontal. The horizontal component is $F \cos \theta$, which overcomes friction. The vertical component is $F \sin \theta$, which reduces the normal reaction.

   Weight $W = 50 \, \text{N}$.

   Normal reaction: $N = W - F \sin \theta$.

   Limiting friction: $f = \mu N$.

   At the threshold of motion: $F \cos \theta = f = \mu N$.

2. Express the condition for minimum force.

   The minimum force to move the body occurs when $\tan \theta = \mu$ (a standard result from optimizing $F = \frac{\mu W}{\cos \theta + \mu \sin \theta}$). At this optimum, the applied force is $F_{\text{min}} = \frac{\mu W}{\sqrt{1+\mu^2}}$.

   We know $F_{\text{min}} = 28.28 \, \text{N}$ and $W = 50 \, \text{N}$. So:

Square both sides:

Note $28.28^2 \approx 800$ (since $28.28 \approx 20\sqrt{2}$, so square is $800$).

Multiply: $800(1+\mu^2) = 2500 \mu^2$ → $800 + 800\mu^2 = 2500\mu^2$ → $800 = 1700\mu^2$ → $\mu^2 = \frac{800}{1700} = \frac{8}{17}$.

So $\mu = \sqrt{\frac{8}{17}} \approx 0.686$.

3. Find the normal reaction and friction. At the optimum angle, $\tan \theta = \mu$, so we can find $\sin \theta$ and $\cos \theta$. If $\tan \theta = \mu$, then $\sin \theta = \frac{\mu}{\sqrt{1+\mu^2}}$ and $\cos \theta = \frac{1}{\sqrt{1+\mu^2}}$. Using $\mu^2 = 8/17$, we have $1+\mu^2 = 25/17$, so $\sqrt{1+\mu^2} = \frac{5}{\sqrt{17}}$. Then:

Now, normal reaction:

Since $28.28 \approx 20\sqrt{2}$, we get:

That gives 34 N, not matching any option. Wait — this suggests the applied force might not be at the optimum angle; instead, the problem likely intends the force to be applied horizontally but with a different interpretation.

The problem gives the initial velocity vector as $\hat{i} + 2\hat{j}$ m/s. This means the horizontal component $u_x = 1$ m/s and the vertical component $u_y = 2$ m/s. We are asked for the Cartesian equation of the projectile's path, with $g = 10$ m/s².

The standard approach: For any projectile launched from the origin, the parametric equations are

$x = u_x t$ and $y = u_y t - \frac{1}{2} g t^2$.

Eliminate $t$ by substituting $t = x / u_x$ into the $y$ equation. This yields a quadratic in $x$ — the path is a parabola.

Let’s work through it step by step.

1. Write the parametric equations

   Horizontal motion (no acceleration): $x = u_x t = 1 \cdot t = t$.

   Vertical motion (constant downward acceleration $g$): $y = u_y t - \frac{1}{2} g t^2 = 2t - \frac{1}{2} \cdot 10 \cdot t^2 = 2t - 5t^2$.

2. Eliminate time $t$

   Since $x = t$, we simply replace $t$ by $x$ in the $y$ equation:

3. Interpret the result …

The key idea is that the rocket is launched with a speed well below Earth’s escape velocity (about 11.2 km/s), so it will rise to a finite height, stop momentarily, and then fall back. Since gravity is not constant with height, we cannot use the simple $mgh$ formula — we must use the full gravitational potential energy expression $-\frac{GMm}{r}$.

Energy conservation is the cleanest route: the sum of kinetic and potential energy at launch equals the sum at the highest point, where kinetic energy is zero.

1. Write the energy conservation equation. At the Earth’s surface ($r = R$), the rocket has kinetic energy $\frac{1}{2}mv^2$ and gravitational potential energy $-\frac{GMm}{R}$. At the maximum height, its distance from Earth’s centre is $r = R + h$, its speed is zero, so only potential energy $-\frac{GMm}{R+h}$ remains.

2. Cancel $m$ and rearrange.

3. Replace $GM$ with $gR^2$. We know $g = \frac{GM}{R^2}$, so $GM = gR^2$. Substituting:

4. Solve for $h$. Multiply both sides by $(R+h)$:

Bring $h$ terms together:

5. Plug in the numbers. $v = 4\ \text{km/s} = 4000\ \text{m/s}$, $R = 6.4 \times 10^6\ \text{m}$, $g = 10\ \text{m/s}^2$. …

The key idea is that rain is falling vertically, but the man is moving horizontally. From his frame of reference, the rain appears to come at him from an angle. To stay dry, he tilts the umbrella into the apparent direction of the rain — that is, along the relative velocity of rain with respect to the man.

Let’s set up the velocities as vectors. Take the vertical direction as positive downward, and the horizontal direction as the man’s direction of motion.

1. Rain velocity relative to ground: The rain falls vertically downward at $6\sqrt{3}$ km/h.

2. Man’s velocity relative to ground: He walks horizontally at $6$ km/h.

3. Relative velocity of rain with respect to the man: This is what the man “sees” — the rain’s velocity in his frame.

The negative $x$-component means the rain appears to come from the front (toward the man’s face, since he walks in the $+\hat{i}$ direction).

4. Angle of this relative velocity with the vertical: The vertical component is $6\sqrt{3}$, the horizontal component is $6$. The angle $\theta$ from the vertical satisfies

Hence

So the rain appears to come down at $30^\circ$ from the vertical, tilted toward the man’s front.

5. How to hold the umbrella: …

The key concept is conservation of mechanical energy for a rigid body that rolls without slipping. Unlike a sliding block (where all potential energy becomes translational kinetic energy), a rolling object must also rotate. The no-slip condition links the rotational speed to the translational speed, so we can express the total kinetic energy in terms of the center-of-mass speed alone.

1. Identify the energy conversion The cylinder starts from rest at height $h$ and ends at the bottom with both translational and rotational kinetic energy. No energy is lost to friction because rolling without slipping means static friction does no work (the point of contact is instantaneously at rest). The height of the incline:

2. Write the conservation of energy equation Initial energy (all potential):

Final energy (translational + rotational):

For a solid cylinder, moment of inertia about its center is

No-slip condition:

3. Substitute and simplify

$m g h = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2$ …

The key idea is that the initial velocity of a projectile is the vector sum of its horizontal and vertical components at the very instant it is launched ($t=0$). The given equations $x = 36t$ and $y = 48t - 4.9t^2$ describe the position over time. To find velocity, we differentiate each component with respect to time.

1. Find the horizontal velocity component The horizontal position is $x = 36t$. Differentiating:

This is constant because there is no horizontal acceleration (we assume air resistance is negligible).

2. Find the vertical velocity component The vertical position is $y = 48t - 4.9t^2$. Differentiating:

At $t=0$, this becomes $v_y = 48\ \text{m/s}$.

3. Combine the components at $t=0$ The initial velocity vector is $(v_x, v_y) = (36, 48)$. Its magnitude (the initial speed) is: $v_0 = \sqrt{36^2 + 48^2} = \sqrt{1296 + 2304} = \sqrt{3600} = 60\ \text{m/s}$ …

The key idea is that the projectile’s motion is given in parametric form: $x(t)$ and $y(t)$. The velocity components are the time derivatives of these functions. At $t=0$, the projectile hasn’t yet been affected by gravity (the vertical component is purely the initial vertical speed), so the initial speed is simply the magnitude of the vector $( \dot{x}(0), \dot{y}(0) )$.

1. Find the horizontal velocity component. The horizontal displacement is $x = 36t$. Differentiate with respect to time:

This is constant because there’s no horizontal acceleration (we assume air resistance is negligible).

2. Find the vertical velocity component. The vertical displacement is $y = 48t - 4.9t^2$. Differentiate:

At $t=0$, this gives $v_y(0) = 48\ \text{m/s}$.

3. Combine to get the initial speed. The initial velocity vector is $(v_x, v_y) = (36, 48)$ m/s. Its magnitude (the initial speed) is: $u = \sqrt{36^2 + 48^2} = \sqrt{1296 + 2304} = \sqrt{3600} = 60\ \text{m/s}.$ …

The key here is to notice where the angle is measured from. Most students are used to measuring angles from the positive x-axis, but this problem explicitly says the angle is $30^\circ$ counterclockwise from the positive y-axis. That changes which trigonometric function gives the y-component.

When a vector makes an angle $\theta$ with the positive y-axis, the y-component is the adjacent side to that angle. So the y-component is $A \cos\theta$, not $A \sin\theta$. The x-component would be $A \sin\theta$ in that case.

Let’s work through it.

1. Identify the given data

   The y-component $A_y = +3.0$ m.

   The angle from the positive y-axis is $\theta = 30^\circ$, counterclockwise.

   The vector lies in the x-y plane.

2. Write the relation for the y-component

   Since $\theta$ is measured from the y-axis,

where $A$ is the magnitude of $\vec{A}$.

3. Substitute the known values

We know $\cos 30^\circ = \frac{\sqrt{3}}{2}$, so

4. Solve for $A$ Multiply both sides by 2:

Then divide by $\sqrt{3}$:

$A = \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$ …

The core idea here is simple: both balls fall freely under gravity from the same height, but they start at different times. The distance between them at the moment the first one lands is just the difference in the distances each has fallen by that instant. Since both accelerate at the same rate, the second ball is always "behind" by the distance it would have covered in the missing 1 s — but because speed increases with time, that gap grows.

Let’s work it through.

1. Time for the first ball to hit ground Use $s = \frac{1}{2} g t^2$ with $s = 500\, \mathrm{m}$ and $g = 10\, \mathrm{m/s^2}$:

So the first ball strikes the ground exactly 10 s after release.

2. Position of the second ball at that instant The second ball is released 1 s after the first, so by the time the first hits ground, the second has been falling for only $10 - 1 = 9$ seconds. Distance fallen by the second ball in 9 s:

That means the second ball is 405 m below the release point — and therefore $500 - 405 = 95\, \mathrm{m}$ above the ground.

3. Separation between the two balls The first ball is on the ground (0 m above ground). The second ball is 95 m above ground. So the vertical distance between them is simply 95 m. …

The key insight here is that the river current and the swimmer’s motion are independent when the swimmer aims straight across. The current only pushes him downstream, but it does nothing to speed up or slow down his crossing. The time to reach the opposite bank depends solely on how fast he moves perpendicular to the banks — which is exactly his still-water speed, because he’s swimming normal to the current.

Let’s walk through it.

1. Set up the situation.

   The river is $1 \ \text{km}$ wide. The swimmer’s speed in still water is $4 \ \text{kmh}^{-1}$, and he strokes perpendicular to the current. The river flows at $3 \ \text{kmh}^{-1}$ parallel to the banks.

2. Separate the velocities.

   Velocity is a vector. The swimmer’s velocity relative to water is $4 \ \text{kmh}^{-1}$ straight across (say along the $y$-axis). The water’s velocity relative to ground is $3 \ \text{kmh}^{-1}$ downstream (along the $x$-axis).

   The swimmer’s velocity relative to ground is the vector sum:

So the $y$-component (crossing direction) is $4 \ \text{kmh}^{-1}$, and the $x$-component (downstream) is $3 \ \text{kmh}^{-1}$.

3. Time to cross depends only on the perpendicular component. The width is $1 \ \text{km}$, and the speed straight across is $4 \ \text{kmh}^{-1}$. $\text{Time} = \frac{\text{distance}}{\text{speed}} = \frac{1 \ \text{km}}{4 \ \text{kmh}^{-1}} = 0.25 \ \text{hours}$ …