Question

A stone is dropped from the top of the tower 500m high into a pond of water at the base of the tower when is the splash heard at the top . G 10 s of the sound 340m/s

Answer 1

Answer:

11.47sec

Explanation:

time required for the stone to reach the bottom of the tower is (2h/g)*1/2 ie, 10 + time taken by the sound to reach the top of the tower = dist travelled/ speed 500/340=1.47

therefore total time req will be 10+1.47=11.47sec

Answer 2

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Height of the tower, s = 500 m

Velocity of sound, v = 340 m sˆ’1

Acceleration due to gravity, g = 10 m sˆ’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, t1

According to the second equation of motion:

$\rm S = ut_1 + 1/2 g t_1^2$

$\rm 500 = 0 \times t_1 + 1/2 \times 10 \times t_1^2$

$\rm t_1^2 = 100$

$\rm t_1 = 10s$

Now, time taken by the sound to reach the top from the base of the tower,

$\rm t_2= 500/340 = 1.47 s$

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

Where, $\rm t= t_1 + t_2 = 10 + 1.47 = 11.47 s.$