Question

A stone dropped from a window reaches the ground in 0.5 seconds (given g = 10 ms2) (i) Calculate the speed just before it hits the ground. (ii) What is the average speed during 0.5? (iii) Calculate the height of window from the ground.

Answer 1

Initial speed (u) = 0  (as the stone is at rest)
Time = 0.5 seconds
Acceleration = 9.8m/s²
i) According to first equation of motion:
v = u + at
v = 0 + 9.8 x 0.5
v = 4.9 m/s
This is the speed just before stone hits the ground
ii) According to third equation of motion :
v² = u² + 2as
(4.9)² = 0 + 2 x 9.8 x s
24.01 = 19.6s
s = 1.225
so the average speed = 1.225/0.5 = 2.45m/s
iii) Height of window from the ground
Height = 1.225m

Answer 2

Initial speed (u) = 0  (as the stone is at rest)

Time = 0.5 seconds

Acceleration = 9.8m/s²

i) According to first equation of motion:

v = u + at

v = 0 + 9.8 x 0.5

v = 4.9 m/s

This is the speed just before stone hits the ground

ii) According to third equation of motion :

v² = u² + 2as

(4.9)² = 0 + 2 x 9.8 x s

24.01 = 19.6s

s = 1.225

so the average speed = 1.225/0.5 = 2.45m/s

iii) Height of window from the ground

Height = 1.225m