Question

Question 13
A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, $g=10m{s}^{-2}$ and speed of sound = $340m{s}^{-1}$.

Answer 1

Height of the tower, s = 500 m
Velocity of sound, $v=340m{s}^{-1}$
Acceleration due to gravity, $g=10m{s}^{-2}$
Initial velocity of the stone, u = 0 (since the stone is initially at rest)
Time taken by the stone to fall to the base of the tower is $t_1$.
According to the second equation of motion:
$S=u{t}_1+\frac{1}{2}g{t}_1^2$
$500=0\times t_1+\frac{1}{2}\times 10\times t_1^2$
$t_1^2=100$
$t_1=10s$
Now, time taken by the sound to reach the top from the base of the tower, $t_2=\frac{500}{340}=1.47s$
Therefore, the splash is heard at the top after time, t.
Where, $t=t_1+t_2=10+1.47=11.47s.$