CBSE Class 9 Answered
Let the stone which is dropped from the terrace be A and the stone thrown upwards be B
Take the vertical downward motion as the positive direction of acceleration of gravity.
Let the stones meet at a depth of x from the top of the building.
Then the height from the ground would be (100m - x)
For the stone A:
initial velocity u = 0m/s
acceleration a= 9.8m/s2
Distance covered s = x
time taken = t
s= ut +(1/2)at2
x= 0+(1/2)x9.8 x t2 => x = 4.9t2
For stone B:
initial velocity u = 25m/s
acceleration a= -9.8m/s2
Distance covered s = 100-x
time taken = t ( as they meet at the same time)
s= ut +(1/2)at2
100- x = 25 t + (1/2) ( - 9.8) t2
=> 100- 4.9t2 = 25 t + (1/2) ( - 9.8) t2
Solving it we get,
t = 4sec
And x = 4.9t2 .Substituting the value of t we get x = 78.4m
Hence the stones meet after 4 sec at a depth of 78.4 metres below the top of the building.