A time `t = 0`, a body of mass `1 kg` is dropped from height `500 m`. At `t = 1 sec`, the body explodes into two parts of equal mass. One of them moves vertically upwards with velocity `5 m//s` just after the explosion. The distance (in `m`) between the parts at time `t = 4 sec` is `15x` Then find `x`?

Answer

6

Explanation

`v = 10 m//s , 1 xx v = (1)/(2) (-5) + (1)/(2) (v'). v'(25), 30 xx 3 = 90`