Question

A stone dropped from the top of a tower of height 300 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 m s–1? (g= 9.8 m s–2)

Answer 1

Height = s = 300m
g = 9.8 m/s²
initial velocity = 0
speed of sound in air = 340m/s
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using newtons second equation of motion

s =  ut + 1/2 gt²
300 = 0 + 1/2 ( 9.8 ) t

300= 4.9t

t² =61.22

t1 = 7.82s
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for sound

we know that velocity = distance /time

therefore 

time = distance / velocity of sound in air

= 300/340

t2 =0.88 s
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for the splash to be heard we need to add the time taked to reach the water and time taken by the sound to reach us

therefore

T = t1 + t2

= 7.82 + 0.88

=8.7s
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after 8.7 s we will hear the splash

Answer 2

let t is the time in which stone strike surface of pond.
if we dropped stone
initial velocity =0
use kinematics formula ,
h=ut+1/2at^2
u=0
h=-h (if we assume reference level is above
g=-9.8 ms^-2

hence,
t=root {2 x 300/9.8} =root(600/9.8) sec
t=7.82 sec
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after strike stone splashes create by water .
time of reach of splashes in top of tower
t"=height of tower/speed of sound
=300/340 sec =0.88 sec

hence,
total time taken=0.88+7.82 =8.7 sec