Question

boy is standing on an open lift moving upwards with speed 10 m/s. The boy throws the ball with speed w.r.t. lift is 24.5 m/s. In how much time the ball returns to the hand of boy? (g = 10 m/s²)


ANS=4.9 SEC
PLS EXPLAIN​

Answer 1

Explanation:

let the constant speed of lift be v' and the initial speed of ball be v (decreasing)

=> v-v'= 24.5 m/s

so the lift and the ball travel the same distance in same time before the boy catches the ball (the lift and ball meet again)

let the distance travelled=s and time taken= t

=> s = vt - 1/2gt^2

and s = v't

equating both equations

=> vt - 1/2gt^2 = v't

=> vt - v't = 1/2gt^2. (g=10m/s^2)

=> (v - v')t = 5t^2

=> v - v' = 5t

=> 24.5 = 5t

therefore, t=4.9s