Numerical Problems Based on Motion under Gravity for Class 11 Physics

Initial velocity, $u = 4.9\text{ ms}^{-1}$ (upwards). Acceleration, $a = -g = -9.8\text{ ms}^{-2}$. At maximum height, final velocity $v = 0$.

(i) Maximum height ($H$):

(ii) Time to reach maximum height ($t$):

(iii) Velocity returning to ground:
By symmetry of motion under gravity, a projectile returns to its launch point with the same magnitude of velocity it was launched with, but in the opposite direction. Therefore, its speed is $4.9\text{ ms}^{-1}$.

(iv) Total time taken to reach the ground ($T$):
Time of ascent equals time of descent. Thus, total time $T = 2t = 2(0.5) = 1\text{ s}$.

Taking the top of the tower as the origin, upward as positive. Net displacement $S = -85\text{ m}$, time $t = 5\text{ s}$, and $a = -10\text{ ms}^{-2}$.

First, find the initial velocity $u$ using $S = ut + \frac{1}{2}at^2$:

(i) Greatest height above the ground:
Maximum height from the top of the tower ($h$) is reached when $v = 0$:

Total height above ground = Height of tower + $h = 85 + 3.2 = 88.2\text{ m}$.

(ii) Velocity with which it reaches the ground:

The magnitude is $42\text{ ms}^{-1}$ downwards.

(iii) Time taken to reach maximum height:

Take origin at the top of the building, upward positive. $u = 9.8\text{ ms}^{-1}$, $a = -9.8\text{ ms}^{-2}$, displacement $S = -39.2\text{ m}$.

(i) Time to reach the ground:
Using $S = ut + \frac{1}{2}at^2$:

Divide the whole equation by 4.9:

Since time cannot be negative, $t = 4\text{ s}$.

(ii) Time to pass the point of projection:
At the point of projection, net displacement $S = 0$.

It passes the origin again at $t = 2\text{ s}$.

(iii) Velocity before striking the ground:

The magnitude of velocity is $29.4\text{ ms}^{-1}$.

Phase 1: Powered flight
Initial velocity $u = 0$, acceleration $a = 10\text{ ms}^{-2}$, time $t_1 = 1\text{ min} = 60\text{ s}$.

Height reached at burnout ($h_1$):

Velocity at burnout ($v_1$):

Phase 2: Free fall under gravity
After fuel finishes, the rocket acts as a projectile. Initial velocity $u_2 = 600\text{ ms}^{-1}$, final velocity $v_2 = 0$, acceleration $a = -g = -9.8\text{ ms}^{-2}$.

Total maximum height ($H$):

When dropped, the packet inherits the balloon’s velocity. Taking upward as positive:
Initial velocity $u = 14\text{ ms}^{-1}$
Displacement $S = -98\text{ m}$ (since it hits the ground below)
Acceleration $a = -9.8\text{ ms}^{-2}$

Velocity on reaching the ground ($v$):

The magnitude is approximately $45.98\text{ ms}^{-1}$ (using $46.0\text{ ms}^{-1}$ is also acceptable depending on required precision).

Time to reach the ground ($t$):

When the stone is dropped, it has the same initial upward velocity as the balloon.
Initial velocity, $u = 16\text{ ms}^{-1}$ (upwards)
Time of flight, $t = 4\text{ s}$
Acceleration, $a = -g = -9.8\text{ ms}^{-2}$

We use the second equation of motion to find the net displacement ($S$) of the stone, which represents the negative of the balloon’s height ($h$) at the time of release.

The negative sign indicates the ground is 14.4 m below the point of release. Therefore, the height of the balloon was $14.4\text{ m}$.