Question

A stone is dropped from the top of a tower 500m high tower into the pond of water at the base of the tower, When is the splash heard at the top?
Given g = 10 ms-² and speed of sound=340ms-¹​

Answer 1

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 = 10 m s-2 and speed of sound = 340 10 m s-1

Height of the tower, s = 500 m

Velocity of sound, v = 340 m/s

Acceleration due to gravity, g = 10 m/s2

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, t1

According to the second equation of motion:

S = ut1 + 1/2gt12

500 = 0 x t1 + 1/2 x 10 x t12

t12 = 100

t1 = 10s

Now, the time is taken by the sound to reach the top from the base of the tower, t2= 500/340 = 1.47 s

Therefore, the splash is heard at the top after time, t

Where, t= t1 + t2 = 10 + 1.47 = 11.47 s.

Answer 2

Answer:

Given :-

• A stone is dropped from the top of a tower 500 m high tower into the pond of water at the base of the tower.
• Acceleration due to gravity is 10 m/s² and the speed of sound is 350 m/s.

To Find :-

• What is the splash heard at the top.

Solution :-

☆ In first case :

Given :

• Distance Covered = 500 m
• Initial Velocity = 0 m/s
• Acceleration due to gravity = 10 m/s²

According to the question by using the formula we get,

$\implies \sf\boxed{\bold{s =\: ut + \dfrac{1}{2} gt^2}}$

$\implies \sf 500 =\: (0)t + \dfrac{1}{2} \times 10 \times t^2$

$\implies \sf 500 =\: 0 \times t + \dfrac{1}{2} \times 10t^2$

$\implies \sf 500 =\: 0 + \dfrac{1}{2} \times 10t^2$

$\implies \sf 500 - 0 =\: \dfrac{1}{2} \times 10t^2$

$\implies \sf 500 =\: \dfrac{1}{2} \times 10t^2$

$\implies \sf 500 \times \dfrac{2}{1} =\: 10t^2$

$\implies \sf \dfrac{1000}{1} =\: 10t^2$

$\implies \sf 1000 =\: 10t^2$

$\implies \sf \dfrac{1000}{10} =\: t^2$

$\implies \sf 100 =\: t^2$

$\implies \sf \sqrt{100} =\: t$

$\implies \sf\bold{t =\: 10\: seconds}$

☆ In second case :

Given :

• Distance Covered = 500 m
• Speed = 340 m/s

According to the question by using the formula we get,

$\implies \sf\boxed{\bold{Time =\: \dfrac{Distance\: Covered}{Speed}}}$

$\implies \sf Time =\: \dfrac{500}{340}$

$\implies \sf\bold{Time =\: 1.47\: seconds}$

Hence, the required splash heard at the top is :

$\dashrightarrow \sf 10 + 1.47$

$\dashrightarrow \sf\bold{\underline{11.47\: seconds}}$

$\therefore$ The splash heard at the top is 11.47 seconds .