Question

A stone is dropped from a height 300m/s and at the same time another stone projected vertically upward with a velocity of 100m/s . Find when and where the stone meet

Answer 1

hlo here is your ans

the hight at which the stone is dropped = 300m
( sorry apne mistake kr rkhi h question m distance ki unit m me hoti h na ki m/s)
g=9.8ms^-1
velocity of the projection is = 100m/s

velocity of projection of some height =0
let the partical be meet at the time t and height h
◆◆◆◆ height covered by the partical from ground
h=ut - 1/2 at^2

h = 100 t -1/2+× 9.8 t^2
h= 100t -4.9t^2
the distance covered by the partical in time t
so it fall down from the rest u =0
300 - h =1/2 gt^2
put value of h in above eq
300- (100t-4.9 t^2)=1/2gt^2

on solving
300-100t
300=100t
t =3
◆◆◆◆◆
height at which the partical meet
h =100t-4.9t^2
h = 100×3 - 4.9 × 3×3
h= 100×3-4.9×9
h=255.9.m