A platform is moving upwards with a constant acceleration of `2 m s^-2`. At time `t = 0`, a boy standing on the platform throws a ball upwards with a relative speed of `8 m s^-1`. At this instant, platform was at the height of `4 m` from the ground and was moving with a speed of `2 m s^-1`. Take `g = 0 m s^-2`. Find
(a) when and where the ball strikes the platform.
(b) the maximum height attained by the ball from the ground.
( c) the maximum distance of the ball from the platform.

Answer

(a) `(10)/(3) s ;`
(b) 9 m
(c ) `(8)/(3) m`.

Explanation

(a) We solve the problem in reference frame of platform,
`vec v_("ball//platform") = 8 hat j`
`vec a_(P//E) = 2 hat j and vec a_(B//E) = -g hat j`
`vec a_(rel) = vec a_(B//P) = -12 hat j`
By `s_(rel) = u_(rel) t + (1)/(2) a_(rel) t^2`
`0 = 8 xx t - (1)/(2) xx 12 t^2 rArr t = (4)/(3) s`
Total time `2 + (4)/(3) = (10)/(3) s`
Displacementof platform in `10//3 s`
=`4 + 2 xx (4)/(3) + (1)/(2) xx 2 xx ((4)/(3))^2 = (76)/(9) m`
(b) `vec v_(B//E) = 10 hat j and vec a_(B//E) = -10 hat j`
By `v^2 = u^2 + 2 as` w.r.t. earth,
`(0)^2 = (10)^2 - 2(10) s_1 rArr s_1 = 5 m`
`H_(max) = 5 + 4 = 9 m`
( c) Also platform frame, `v^2 = u^2 + 2as`
`(0)^2 = (8)^2 + 2(-12) s rArr s = (8)/(3) m`.