Question

A ball is dropped from a bridge at a height of $176.4\mathrm{m}$ over a river. After $2\mathrm{s}$, a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water simultaneously?

• A. $2.45{\mathrm{ms}}^{-1}$
• B. $49{\mathrm{ms}}^{-1}$
• C. $25.7{\mathrm{ms}}^{-1}$
• D. $24.5{\mathrm{ms}}^{-1}$

Answer 1

The correct option is D

$24.5{\mathrm{ms}}^{-1}$

Step 1: Given data

Let us consider,

Initial velocity of first ball$={\mathrm{u}}_1=0$

Initial velocity of second ball$={\mathrm{u}}_2$

Height of a ball, $\mathrm{S}=176.4\mathrm{m}$

Time interval between first and second ball$=2\mathrm{s}$

Step 2: Calculating initial velocity of second ball.

Using third equation of motion for first ball, we get

$$\begin{gathered} \mathrm{S}={\mathrm{u}}_1\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ \Rightarrow 176.4=\frac{1}{2}{\mathrm{at}}^2\because {\mathrm{u}}_1=0 \\ \Rightarrow 176.4=\frac{1}{2}\times 9.8\times {\mathrm{t}}^2 \\ \Rightarrow \mathrm{t}=6\mathrm{s} \end{gathered}$$

Using third equation of motion for second ball, we get

$\mathrm{S}={\mathrm{u}}_2(\mathrm{t}-2)+\frac{1}{2}\mathrm{a}{(\mathrm{t}-2)}^2\left(1\right)$

Substitute $\mathrm{t}=6\mathrm{s}$ in (1) equation we get,

$$\begin{gathered} \mathrm{S}={\mathrm{u}}_2(6-2)+\frac{1}{2}\times 9.8\times {(6-2)}^2 \\ \Rightarrow 176.4=4{\mathrm{u}}_2+78.4 \\ \Rightarrow 4{\mathrm{u}}_2=98 \\ \Rightarrow {\mathrm{u}}_2=\frac{98}{4}=24.5{\mathrm{ms}}^{-1} \end{gathered}$$

Hence, option (D) is correct answer.