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.
According to Figure the height of tower is 100m, Assuming the two stones meet at point P which is at a height x from the ground, from this we get that distance between Point P and the top of tower is (100-x)
(i)Firstly, for the stone falling from top of tower:
Height, h = (100 - x)
Initial velocity, u = 0Time, t =?
g = 10 m/s2 (As the stone is falling down)We know that,
Putting the values in above equation
(100 – x) = 0 × t +1/2× 10 × t2
(100 - x) = 5t2 (i)(ii)Now, for stone projected vertically upwards:
Height, h = xInitial velocity, u = 25 m/s
Time, t =?g = -10 m/s2 (As the stone is going up)
We know that,
Putting the values in above equation
x= 25×t+1/2×(-10)×t2
x = 25t – 5t2 (ii)
Now, by adding (i) and (ii), we get:
100 – x + x = 5t2 + 25 t – 5t2
100 = 25 tt = 
Therefore, the 2 stones will meet after a time of 4 seconds.
Now, putting the value of t in (i), we get:100 – x = 5× (4)2
100 – x = 5× 16100 – x = 80
x = 20m
Therefore, the two stones will meet at a height of 20 m above the ground after 4 seconds
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