Question

An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?

Answer 1

The acceleration due to gravity is -9.8m/s2.
Make use of the equations of motions:
SUVAT-- s-displacement, u-initial velocity, v-final velocity,a- acceleration, t-timev=u + atv2 = u2 + 2ass=1/2(u+v)ts=ut + 1/2at2
The values for the first object:
t=2 seconds
a=9.8 m/s2
u= 0m/s
s=???

s= ut +1/2at2  ut=0x2=0        s= 1/2 x 9.8 x 2^2
                                              s=   9.8x2= 19.6
 Their height of first object after 2 seconds= 150-19.6 = 130.4m

Values for the second object:
 a=9.8m/s2
 t=2 secs
 u=0m/s
 s=???

s= ut + 1/2at^2
s=1/2 x 9.8 x 2^2 =   9.8 x 2= 19.6
therefore height of second object after 2 seconds= 100-19.6 
                                                                              =80.4m 
a)Therefore difference in height after 2 seconds= 130.4- 80.4 = 50meters

b)How does height vary with time.

When you calculate the height after the next 2 seconds, the difference is still 50m.

Therefore the difference in height is 50 meters after every 2 seconds
=50m/2seconds  25m/s

That means a difference in height at the rate of 25 meters per second.