Question

A stone is dropped into water from a bridge $44.1\text{m}$ above the water. Another stone is thrown vertically downward $1\text{sec}$ later. Both strike the water simultaneously. What was the initial speed of the second stone.

• A. $12.25\text{m/s}$
• B. $14.75\text{m/s}$
• C. $16.23\text{m/s}$
• D. $17.15\text{m/s}$

Answer 1

The correct option is A $12.25\text{m/s}$

Taking dropping point as origin and downward direction as negative and upward position as positive, we have from equation of motion
$S=ut+\frac{1}{2}a{t}^2$
For first stone, since it is dropped, we get
$-44.1=0-\frac{1}{2}\times 9.8\times t^2$
$\Rightarrow t=3$ sec
Now, for second stone, we have
$t=3–1=2\text{sec}$
So, from equation of motion we have
$S=ut+\frac{1}{2}a{t}^2$
$\Rightarrow -44.1=-u\left(2\right)-\frac{1}{2}\times 9.8\times 2^2$
$\Rightarrow 2u=44.1–19.6$
$\Rightarrow u=12.25\text{m/s}$
Hence, the initial speed of second stone is $12.25\text{m/s}$