Question

A ball is projected with a speed $v_0$ at an angle of $37°$ with the horizontal, from the top of a cliff as shown in the figure. The ball has been thrown with the aim so that it just hits directly the edge of the cliff, but the ball just misses the target and falls on ground as shown in figure. [Take $\sin 37°=\frac{3}{5};\cos 37°=\frac{4}{5};g=10m{s}^{-2}$.]

Based on above information, answer the following question:
The time taken by ball to reach ground is

• A. $4.7s$
• B. $3.8s$
• C. $4.1s$
• D. $5.0s$

Answer 1

The correct option is A

$4.7s$

Step 1: Given data

The range of the projectile motion till it hits the cliff, $R=60m$

The angle of projection, $\theta =37°$

The initial velocity of the ball is $v_0$

The acceleration due to gravity, $g=10m{s}^{-2}$

Step 2: Find the initial velocity of the ball

The formula for the range of the projectile motion, $R=\frac{u^2\sin 2\theta }{g}$

$$\begin{gathered} \Rightarrow 60=\frac{{v_0}^2\times 2\sin 37°\cos 37°}{10}\left[\because \sin 2\theta =2\sin \theta \cos \theta \right] \\ \Rightarrow 60=\frac{{v_0}^2\times 2\times {\displaystyle \frac{3}{5}}\times {\displaystyle \frac{4}{5}}}{10} \\ \Rightarrow {v_0}^2=\frac{60\times 10\times 5\times 5}{2\times 3\times 4} \\ \Rightarrow {v_0}^2=625 \\ \Rightarrow v_0=\sqrt{625} \\ \Rightarrow v_0=25m{s}^{-1} \end{gathered}$$

Step 3: Find the time of flight of the ball

The formula for the time of flight in the projectile motion, $T=\frac{2v_o\sin \theta }{g}$

$$\begin{gathered} \Rightarrow T=\frac{2\times 25\times {\displaystyle \frac{3}{5}}}{10} \\ \Rightarrow T=4.7s \end{gathered}$$

The time taken by the ball to reach the ground is $4.7s$.

Hence, option A is correct.