Question

a boy dropped a stone in a well 45 m deep. if speed of sound is 340 m/s, then after how much time ,he will hear the splash? take g = 10m/s^2.

Answer 1

Time, after which splash is heard =  time acquired by the stone to reach the                                                      surface of water in a well + time taken by                                                  sound of splash to reach the top of tower.

i). Here,           u = 0
                        s = 500m
                        g = 10 m/s²

using,            s = ut + \frac{1}{2} gt², we have
                 
                     45 = 0 + $\frac{1}{2}$ x 10 t² 
         
                     45   = 5t²
                       t² = 45 / 5
                     t = √9
                     t = 3 s

Therefore time taken by the stone to reach the surface of water = 3 s

ii). Here,            s = 45 m,   v = 340 m/s
           
         using,  s = ut, we get
               t = s/u
                = 45 / 340
                = 0.13s

Thus time taken by the sound to reach the top of water = 0.13 s
    
   ∴ Time, after which splash is heard
                          = 3s + 0.13s
                          = 3.13s

Answer 2

Explanation:

Time, after which splash is heard = time acquired by the stone to reach the surface of water in a well + time taken by sound of splash to reach the top of tower.

i). Here, u = 0

s = 500m

g = 10 m/s²

using, s = ut + \frac{1}{2} gt², we have

45 = 0 + \frac{1}{2}

2

1

x 10 t²

45 = 5t²

t² = 45 / 5

t = √9

t = 3 s

Therefore time taken by the stone to reach the surface of water = 3 s

ii). Here, s = 45 m, v = 340 m/s

using, s = ut, we get

t = s/u

= 45 / 340

= 0.13s

Thus time taken by the sound to reach the top of water = 0.13 s

∴ Time, after which splash is heard

= 3s + 0.13s

= 3.13s