Question

a ball is thrown downward with a speed of 20m/s from top of a building 150 m high and simultaneously another ball is thrown vertically upward with a speed of 30 m/s from the foot of a building then the time after which both the ball will meet ??

Answer 1

it can be solved easily with help of relative concept.

actually, time taken to meet = relative seperation/relative velocity.

here, velocity of first ball (thrown downward) , $v_1=20m/s$

velocity of 2nd ball ( thrown upward), $v_2$ = 30m/s

seperation between balls , s = 150m

time taken to meet , t = relative seperation/relative velocity

= $\frac{s}{v_2-v_1}$

= 150/(30 - 20)

= 150/10 = 15sec

hence, 15 sec after which both the ball will meet.

Answer 2

Answer:3 sec

Explanation:s1=20t+5t^2

s2=30t-5t^2

s1+s2=150

150=50t

t=3 sec