Question

A ball A is projected from O with an initial velocity $v_0=7m{s}^{-1}$ in a direction $37°$ above the horizontal. Another ball B, $3m$ from O on a line $37°$ above the horizontal is released from rest at the instant A starts, as shown in figure.

[Take $\sin 37°=\frac{3}{7};\cos 37°=\frac{4}{5};g=9.8m{s}^{-2}$]

Based on above information, answer the following questions:

How far will B have fallen when it is hit by A?

• A. $\frac{1}{10}m$
• B. $\frac{9}{10}m$
• C. $\frac{196}{25}m$
• D. $\frac{81}{10}m$

Answer 1

The correct option is B

$\frac{9}{10}m$

Step 1: Given data

The initial velocity of ball $A$, $v_0=7m{s}^{-1}$

Therefore, the vertical component of the initial velocity of ball $A$, $v_y=7\sin 37°m{s}^{-1}$

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

The distance of the ball $B$ from the origin is$3m$.

As the ball $B$ is dropped initially, its initial velocity is $u_B=0m{s}^{-1}$

(Note: The downward direction is considered a negative direction here.)

Step 2: Calculate the time of the collision

From the above figure in the right-angled triangle$∆BCA$:

$$\begin{gathered} \cos 37°=\frac{OC}{AB} \\ \Rightarrow OC=AB\times \cos 37° \\ \Rightarrow OC=3\times \frac{4}{5} \\ \Rightarrow OC=\frac{12}{5}m \end{gathered}$$

As the ball $B$ is moving in a vertically downward motion, it means that in order to collide with $B$ the horizontal displacement of $A$ will be $OC$ at time of collision $t$.

Using the third equation of motion under the gravity in the horizontal direction ${\mathit{s}}_{\mathbf{x}}\mathbf{=}{\mathit{u}}_{\mathbf{x}}\mathit{t}$ we get,

$$\begin{gathered} \Rightarrow OC=v_0\cos 37°\times t \\ \Rightarrow \frac{12}{5}=7\times \frac{4}{5}\times t \\ \Rightarrow t=\frac{12\times 5}{5\times 7\times 4} \\ \Rightarrow t=\frac{3}{7}s \end{gathered}$$

Step 3: Calculate the distance traveled by ball B before hitting A

The direction of motion of ball B and direction of gravity is vertically downward.

Using the third equation of motion under the gravity for ball B is ${\mathit{s}}_{\mathbf{y}}\mathbf{=}{\mathit{u}}_{\mathbf{y}}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{g}}_{\mathbf{y}}{\mathit{t}}^{\mathbf{2}}$ we get,

For ball B at the time of collision:

$$\begin{gathered} \Rightarrow s_B=0\left(t\right)+\frac{1}{2}g{t}^2 \\ \Rightarrow s_B=\frac{g{t}^2}{2} \end{gathered}$$

$$\begin{gathered} \Rightarrow s_B=\frac{9.8\times {\left({\displaystyle \frac{3}{7}}\right)}^2}{2} \\ \Rightarrow s_B=\frac{9.8\times {\displaystyle \frac{9}{49}}}{2} \\ \Rightarrow s_B=\frac{1.8}{2} \\ \Rightarrow s_B=0.9 \\ \Rightarrow s_B=\frac{9}{10}m \end{gathered}$$

Hence, option B is the correct option.