A bus starts from rest accelerating uniformly with `4ms^(-1)` at t =10s a stone is dropped out of a window of the bus 2m high what are the (i) magnitude of velocity and (ii) accelearatio of the stone at 10.2 s take g=10 `ms^(-2)`

Explanation

Given
Acceleration `=4ms^(-2)`
Time =- 10s
(i) horizontal velocity of the bus or the stone at t= 10 is
`v_(x) = u + at = 0 + 4 xx 10`
for vertical motion of the stone
`u= 0 a=g = 10 ms^(-2) t =10.2 -10 =0.2s`
`v_(y) =0+ 10 xx 0.2 2ms^(-2)`
Magnitude of teh resultant velocity of the stone is
`v= sqrt( v^(2) + v^(2)) = sqrt(40^(2) +2^(2))`
(ii) After teh stone is dropped its acceleration along horizontal is zero . it has only a vertical acceleration of `10 ms^(-2)`