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. Calculate when and where the two stones will meet.
Here, h = 100 m.
Let the two stones meet after t seconds at a point P which is at a height x above the ground as shown in figure.
For stone 1,
u = 0, h = (100 - x)m
a = g = 9.8 m/s2
From s = \(ut + \frac{1}{2}at^2\)
(100 - x) = \(0 + \frac{1}{2}\times 9.8t^2 = 4.9t^2\) .....(i)
For stone 2,
u = 25 m/s, h = x, a = - g = -9.8 m/s2
From s = \(ut + \frac{1}{2}at^2\)
x = \(25t+ \frac{1}{2}(9.8)t^2 = 25t-4.9t^2\) ....(ii)
100 - x + x = 25t
= t = \(\frac{100}{25} = 4s\)
From equation (i),
100 - x = 4.9 x \((4)^2\) = 78.4
x = 100 - 78.4 = 21.6 m