一題PHY, 計高度

Take the downward direction be positive.

Initial velocity of the stone, u = 0

Acceleration of the stone due to gravity, g = 10 ms^-2

By s = ut + 1/2 at²

s = (0)t + 1/2 (10)t²

s = 5t² ───── (1)


The sound travelled at a uniform velocity, v = 330 ms^-1

time travelled by the sound wave = 5.38 - t

So, s = (330)(5.38 - t)

s = 1775.4 - 330t ───── (2)


(1) = (2):

5t² = 1775.4 - 330t

5t² + 330t - 1775.4 = 0

t = 5.00 or -71.0 (rejected)


Sub t = 5.00 into (1)

s = 5(5.00)²

s = 125 m


Therefore, the height of the cliff is 125 m.

Considering the stone falling vertically,
s = ut + 1/2at^2
s = 5t^2 .............(1)
[since u = 0ms-1 and a = 10ms-2]

Considering the impact sound,
s = v't' = 330 x (5.38 - t) .......(2)
[v' = velocity of sound and t' = time for sound to travel after impact = (5.38 - t)]

(1) = (2),
5t^2 = 330 x (5.38 - t)
5t^2 + 330t - 330x5.38 = 0
t^2 + 66t - 355.08 = 0
t = 5.00s or -71s(rejected) .......(3) [solving quadratic equation]

Sub (3) into (1)
s = 5x (5.00)^2 = 125m