Question

a ball is thrown up with velocity 20m/s relative to ground in open lift moving up with speed 10m/s at time of throwing and has an acceleration of2m/s upward ball meets lift after​

Answer 1

Hii Dear,

◆ Answer -

t = 1.67 s

● Explanation -

For a ball,

u = 20 m/s, a = -g = -10 m/s^2

Let the ball travels distance s till it touches lift.

s = ut + 1/2 at^2

s = 20×t + 1/2 × (-10) × t^2

s = 20t - 5t^2 ...(1)

For a lift,

u = 10 m/s, a = 2 m/s^2

Distance s travelled by lift till it touches the ball is -

s = ut + 1/2 at^2

s = 10×t + 1/2 × 2 × t^2

s = 10t + t^2 ...(2)

Equating (1) & (2),

20t - 5t^2 = 10t + t^2

20t - 10t = t^2 + 5t^2

10t = 6t^2

t = 5/3 s

t = 1.67 s

Therefore, ball will meet the lift after 1.67 s.

Thanks dear...