Question

A ball is dropped from the top of a building of hight 100m at time t=0.At same instant another ball is thrown upwards with speed 50m/s from the bottom of the building the time at which ball will meet kis​

Answer 1

answer : 2 sec

explanation : let time taken to meet = t sec

distance covered by first ball( dropped from the top of building) , $s_1$ = 0 + 1/2 gt² [ as initial velocity of dropped ball equals zero]

distance covered by second ball (thrown upward) , $s_2$ = 50t - 1/2 gt²

balls will meet when $s_1+s_2$ = 100 m

or, 1/gt² + 50t - 1/2 gt² = 100

or, 50t = 100

or, t = 100/50 = 2sec.

shortcut :

time taken = relative seperation/relative velocity

= {0 - (-100)}(50 - 0)

= 100/50 = 2 sec