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Question

# Question 17 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.

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Solution

## Let t be the time at which two stones meet and let h′ be their height from the ground. Given that height of the tower, H = 100m First consider the stone which falls from the top of the tower. Distance covered by this stone in time t can be calculated using the second equation of motion. h=ut+12gt2 Since initial velocity u = 0 so we get h=12gt2……(1) The distance covered by the stone that is thrown vertically upwards is h′=ut−12gt2 In this case, initial velocity is 25 m/s. So, h′=25t−12gt2……(2) Adding equations (1) and (2) we get, h + h' = 25t But h + h' = H = 100 m So, 100 = 25 t or, t = 4 s Putting value in equation (2), h′=25×4−12×9.8×(4)2=100−78.4=21.6 m They will meet at a height of 21.6 m from ground.

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