Question

A stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward one second later. Both strike the water simultaneously, then the initial speed of the second stone is

• A. $12.25m{s}^{-1}$
• B. $14.75m{s}^{-1}$
• C. $16.23m{s}^{-1}$
• D. $17.15m{s}^{-1}$

Answer 1

The correct option is A $12.25m{s}^{-1}$

$\text{Step 1: Time taken (t) by stone 1}$

(Taking downward positive)

Since acceleration is constant, therefore applying equation of motion

$h=ut+\frac{1}{2}g{t}^2$

$\Rightarrow$ $44.1=0+\frac{1}{2}\left(9.8\right)t^2$

$\Rightarrow$ $t=3s$

$\text{Step 2: Initial Speed Calculation}$

Let the initial speed of stone $2$ be $u$

Both the stones strike the water simultaneously but the stone $2$ is thrown $1s$ later which means it takes $1s$ less as compared to stone $1$ to cover the same distance.

Therefore, time taken by stone $2$, $t_1=t-1$ $=3-1s=2s$

Since acceleration is constant

$\therefore$ Applying equation of motion (Taking downward positive)

$h=u{t}_1+\frac{1}{2}g{t}_1^2$

$\Rightarrow 44.1=u\times 2+\frac{1}{2}\times 9.8\times (2{)}^2$

$\Rightarrow u=12.25m/s$

Hence option $A$ is correct.