Question

An object is dropped from rest at a height of 150 m and another object is dropped from rest at a height of100 m.What is the difference in their height after 2 sec if both the object drop with same acceleration?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.

Answer 2

If a an object is falling under the gravity the the acceleration between the both object is the same as one change it speed another also change the speed with the same acceleration so at the start of the fall the distance between both object is 50m and after 2 second it will remain 50m