A body is allowed to fall from the top of a building 200 m high and at the same time another
body is projected vertically upwards from the ground with a velocity of 51 m/s. The two stones
will meet above the ground at a height of (given g = 10 ms)
a) 100 m
b) 120 m
c) 140 m
d) 160 m
Answers
1. one must find out the time taken by the body sent upward to meet the top body (as if the body was not dropped)
2.so if you divide the height 200m by the velocity 50m/s then one gets 4 sec.
3.if you do all equations as done by another responder for accelerated motion ultimately you will get the same answer.
4.you can ask why you are getting the correct time interval even if g was put off.it so happens that the acceleration produced each sec to the top body (dropped)is also a deceleration produced to the velocity of bottom particle, so net effect of g gets adjusted.
5.now to find where they will meet -the top one will travel a distance in 4 sec d= (1/2) g . t^2 = (1/2) x 9.8 x 16 = 78.4 meters from the top ; so from the bottom the distance would be d’= 200–78.4 = 121.6 m