Numerical Problems Based on Projectile Fired Horizontally for Class 11 Physics

Initial horizontal velocity $u_x = 100\text{ ms}^{-1}$. Initial vertical velocity $u_y = 0$. Height $h = 1000\text{ m}$. Take $g = 9.8\text{ ms}^{-2}$.

(i) Time taken to reach the ground ($t$):
Using vertical motion equation $h = u_y t + \frac{1}{2}gt^2$:

(ii) Velocity hitting the ground ($v$):
Horizontal velocity remains constant: $v_x = 100\text{ ms}^{-1}$.
Final vertical velocity $v_y = u_y + gt = 0 + 9.8(14.28) \approx 139.94\text{ ms}^{-1}$.
Resultant velocity magnitude:

Angle $\beta$ with the horizontal:

(iii) Distance of the target (Horizontal range $R$):

Height $h = 19.6\text{ m}$. Initial vertical velocity $u_y = 0$. $g = 9.8\text{ ms}^{-2}$.

Time to strike the ground ($t$):

Initial velocity ($u_x$):
The line joining the start to end points forms a right triangle with the horizontal range $R$ and height $h$. Given the angle is $45^\circ$:

So, the horizontal range $R = 19.6\text{ m}$.
Since horizontal motion has constant velocity, $R = u_x \times t$:

Time $t = 2\text{ s}$. Angle at which it strikes the ground $\theta = 45^\circ$.

Height of the tower ($h$):
Using $h = u_yt + \frac{1}{2}gt^2$ with $u_y = 0$:

Speed with which it was thrown ($u_x$):
The vertical velocity when it hits the ground is $v_y = u_y + gt = 0 + 9.8(2) = 19.6\text{ ms}^{-1}$.
It strikes the ground at an angle of $45^\circ$, which means the horizontal and vertical components of the final velocity are equal in magnitude:

Since horizontal velocity is constant, the initial speed $u_x = v_x = 19.6\text{ ms}^{-1}$.

Horizontal distance $R = 200\text{ m}$.
Vertical drop required, $h = 540\text{ m} – 50\text{ m} = 490\text{ m}$.
Take $g = 9.8\text{ ms}^{-2}$.

First, find the time of flight ($t$) for this vertical drop:

Now, calculate the required horizontal speed ($u_x$):

When dropped, the stone inherits the horizontal velocity of the bus.
$u_x = 60\text{ kmh}^{-1} = 60 \times \frac{5}{18} = \frac{50}{3}\text{ ms}^{-1} \approx 16.67\text{ ms}^{-1}$.
Vertical height $h = 1.96\text{ m}$. Initial vertical velocity $u_y = 0$. Take $g = 9.8\text{ ms}^{-2}$.

Time of flight ($t$):

Horizontal distance ($R$):

Initial horizontal velocity $u_x = 600\text{ kmh}^{-1} = 600 \times \frac{5}{18} = \frac{3000}{18} = \frac{500}{3}\text{ ms}^{-1}$.
Height $h = 1960\text{ m}$. Take $g = 9.8\text{ ms}^{-2}$.

Time of flight ($t$):

The distance AB is the horizontal range $R$:

Initial horizontal velocity $u_x = 270\text{ kmh}^{-1} = 270 \times \frac{5}{18} = 15 \times 5 = 75\text{ ms}^{-1}$.
Height $h = 176.4\text{ m}$. Take $g = 9.8\text{ ms}^{-2}$.

Time of flight ($t$):

The horizontal distance the bag will travel before hitting the ground is the required distance: