A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25m/s . Calculate when and where the two stones will meet .
Dear Student,
Suppose the two stones meet at a height ‘x’ from the ground.
The stone that is dropped from above will cover (100-x) before meeting the stone that is thrown from ground.
For the stone that is thrown from above the time taken to cover (100-x) is say, t.
(100-x) = (0)(t) + ½ (10)t2
=> 100 – x = 5t2 …………….(1)
For the stone that is thrown from the ground will cover distance ‘x’ in the same time ‘t’.
So, x = 25t – ½ (10)t2
=> x = 25t -5t2 ………………(2)
(1) + (2) => 100 = 25t
=> t = 4 s
Thus, the two stones will meet 4 s from the time of projection.
Using (2),
x = (25)(4) – (5)(42)
=> x = 100 – 80
=> x = 20 m
The stones meet 20 m above the ground.
Regards