Question

A stone is dropped from the top of a tower 500m highin to a pond of water at the base of tower.when is the splash heard at the top

Answer 1

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$\Huge{\boxed{Question}}$

=> A stone is dropped from the top of a tower 500m high in to a pond of water at the base of tower.when is the splash heard at the top?

[ Given - g = $10ms^{-2}$ and speed of sound - $340ms^{-1}$ }

$\italic{\sf[Solution}}$

Given :

• Distance = 500m
• intial velocity = 0
• g = $10ms^{-2}$

• time = ?  

Now we find intial time :

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

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

                                        $t^{2}$ = $\frac{500}{5}$

                                     $t^{2}$ = 100

                                       t = 10 second  ................1]    

Time taken by sound to travel from base of the tower to its top

Now we find final time :

∴ $t^{,}$ = $\frac{Distance }{speed of sound } = \frac{500 }{340ms^{-1} }$

=> 1.47 second .................2]

Total time after which the splash is heard

=> first + second = 10 + 1.47 => 11.47 second  answer

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