Question

a stone is dropped from the top of a tower of the tower 500 m high into a Pound of water at the base of the tower. when is the splash hot at the top? given g = 10m/s^2 and speed of sound = 340 m s^-1​

Answer 1

❒Solution

time taken by the stone to reach the bottom of the tower....

formula used

$s = ut + \frac{1}{2} at {}^{2}$

s=500 m

a=g=10m /s²

u=0

$500 = 0 \times t + \frac{1}{2} \times 10 \times t {}^{2} \\ = > t = \sqrt{100} = 10 \: \: sec$

now. ....

time taken by the sound to travel from the base to the tip of the tower....

$t(1) = \frac{distance}{speed \: of \: sound} = \frac{500}{340} \: sec = 1.47 \: \: sec$

therefore...total time after which the splash is

heard

=t+t(1)=(10+1.47)sec=11.47 sec

hope this helps you......

Answer 2

Answer:

s = 500m

u= 0

t= ?

g = 10 m s^-2

s = ut+1/2gt^2

500= 5 ×t^2

500= 5 ×t^2

t^2= 500÷5

t^2= 100

t=√100

t = 10

Now the speed of sound

speed of sound = distance / time

340 = 500/time

time = 500/340s

1.47 s

time after splash is 10s+1.47s

= 11.47 s