A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Where are when will the two stones meet? Chapter - Motion & Kinematic Formulas Class IX
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let the two stone meet at C after time t .
let AC = x
BC = 100-x
Taking vertical downward motion of stone .
u= 0 , a = 9.8m/s^2 ,S =x , t= t
As
S = ut + 1/2 at^2
x= 0 + 1/2 x 9.8 x t^2
x= 4.9 t^2 -------------(1)
Taking vertical upward motion of the stone thrown from B
u = 25 m/s
a= 9.8 m/s
S = ut + 1/2 at^2
100-x = 25t + 1/2 ( -9.8 )t^2
= 25t- 4.9 t^2 ----------------(2)
Add (1) and (2)
100 = 25 t
t= 4s
putting value in (1)
x= 4.9 x 16
= 78.4 m
in 4s and at a distance of 78.4 m below from top .