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 question:

After how much time from the instant of projection of A, the two balls collide?

• A. $1/7s$
• B. $3/7s$
• C. $2/5s$
• D. $9/7s$

Answer 1

The correct option is B

$3/7s$

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 perpendicular distance of the ball $\mathit{B}$ from the ground

In the right-angled triangle $BOC$, using the formula $\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }\mathbf{=}\frac{\mathbf{P}\mathbf{e}\mathbf{r}\mathbf{p}\mathbf{e}\mathbf{n}\mathbf{d}\mathbf{i}\mathbf{c}\mathbf{u}\mathbf{l}\mathbf{a}\mathbf{r}}{\mathbf{H}\mathbf{y}\mathbf{p}\mathbf{o}\mathbf{t}\mathbf{e}\mathbf{n}\mathbf{u}\mathbf{s}\mathbf{e}}$.

$$\begin{gathered} \Rightarrow \sin 37°=\frac{BC}{OB} \\ \Rightarrow \frac{3}{7}=\frac{BC}{OB} \\ \Rightarrow BC=3\times \frac{3}{7} \\ \Rightarrow BC=\frac{9}{7} \end{gathered}$$

Step 3: Calculate the distance traveled by both the balls till the time of the collision

Let A and B collide after time $t$.

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

For ball A:

$$\begin{gathered} \Rightarrow s_A=7\sin 37°\times t-\frac{1}{2}g{t}^2 \\ \Rightarrow s_A=7\times \frac{3}{7}\times t-\frac{1}{2}g{t}^2 \\ \Rightarrow s_A=3t-\frac{g{t}^2}{2} \end{gathered}$$

For ball B:

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

Step 4: Calculate the time of the collision

As the distance of the two balls will be the same from the ground at the time of collision, we get: $\mathit{B}\mathit{C}\mathbf{=}{\mathit{s}}_{\mathbf{A}}\mathbf{-}{\mathit{s}}_{\mathbf{B}}$

$$\begin{gathered} \Rightarrow \frac{9}{7}=3t-\frac{g{t}^2}{2}-\left(-\frac{g{t}^2}{2}\right) \\ \Rightarrow \frac{9}{7}=3t-\frac{g{t}^2}{2}+\frac{g{t}^2}{2} \\ \Rightarrow \frac{9}{7}=3t \\ \Rightarrow t=\frac{3}{7}s \end{gathered}$$

Hence, option B is the correct option.