Question

A stone is dropped from the top of a tower 500 m high into a pond of water at the base of tower . when is splash heard at the top ? ​

Answer 1

Answer:

Thats very simple bro

       given height,s=500m

                g=10

       speed of sound =340

 on analysing we get that  we have to find time period

first analyze the data and apply different formulas for it then see which formula suits the best.

         the formula that should be used here is

                                   s=ut+1/2at2

                          u=0,

                  500=0+1/2x10xtxt

we get t=10 sec

 now the time for the resound=500/340=1.47 sec

 total time=10+1.47=11.47 sec

عید مبارک ہو

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