In: Physics
A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 2.5 s later. The speed of sound in air is 343 m/s. How high is the cliff?
Solution:
Let t1 be the time taken by the rock to touch the water and.
t2 be the time taken by the sound to reach us.
Given that t1 + t2 = 2.5
s -------------(1)
Velocity of sound v = 343 m/s
Let S be the distance taveled by the rock.
t2 = S/v = S/343 ........... (2)
from eq. of motion, S =
(1/2)g(t1)2 (initial speed
u =0)
t2 =
[(1/2)(9.8 m/s2)(t1)2]/(343 m/s) =
0.014286(t1)2 ..........(3)
substitute the eq (3) in eq (1), we get
0.014286(t1)2 +
t1 - 2.5 = 0
On solving the above quadratic equation we get
t1 = 2.416 s
Hence distance covered by the rock is
S = (1/2)(9.8 m/s2)(2.416 s)2
S = 28.63 m