Question

A stone dropped from top of a tower of height 280 m high, splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 ms–1 ? (g = 9.8 ms–2 )

Answer 1

Answer:

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:

I hope it will help you.

Thank you!!!!