A man drops a ball from a height of 180 m. How long will it take for the ball to reach the ground and what is the veloicity of the ball on impact.
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A ball is dropped from a height of 180 m. The ball moves down as it is attracted towards the Earth due to the gravitational force. The acceleration due to this force can be considered constant and equal to 9.8 m/s^2.
The initial velocity of the ball is 0 m/s. The distance traveled by an object in time t is given by S = u*t + (1/2)*a*t^2 where u is the initial velocity and a is the acceleration. If v is the velocity after traveling a distance s, the relation v^2 - u^2 = 2*a*s applied.
Using the data provided in the problem, 180 = 0*t + (1/2)*9.8*t^2
=> t = `sqrt(180/4.9)` = `(30*sqrt 2)/7` s
v^2 - 0 = 2*9.8*180
=> v = `42*sqrt 2` m/s
The distance taken by the ball to fall 180 m is `(30*sqrt 2)/7` s and the velocity on impact is `42*sqrt 2` m/s
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