A bus starts from rest accelerating unifromly with `4 ms^(-2)` . At `t = 10 s` ,a stone is dropped from the window of the bus 2 m high . If `g = 10 m//s^(2)` , what are the magnitude of velocity and acceleration of the stone at 10 .2 s ?

Explanation

Horizontal velocity of bus/stone at
`t = 10 s`
`upsilon = u + at = 0 +4 xx 10 = 40 ms^(-1)`, for vertical motion of the stone ,
`u = 0 , a = g = 10 m//s^(2) , t = 10 .2 - 10 .0 = 0.2 s`
`upsilon_(y) = u + at =0 + 10 xx 0.2 = m//s`
Horizontal vel . of stone = velocity at t = 10 s when it is dropped
`upsilon_(x) = 40 m//s` . Resultant velocity of stone `=sqrt(upsilon^(2)_(x)+upsilon^(2))`
`= sqrt(40^(2) +2^(2))=40.04 ms^(-1)`
After the stone is dropped , its acceleration along the horizontal is zero . The stone has only vertical accleration `= 10 m//s^(2)`