Question:
Jane drops a football from the roof onto the ground below. The ball weighs 0.8kg and the distance the ball falls is 5m. Assuming there are no external forces acting on the ball, what speed will the ball be travelling at just before it hits the ground?
Methods:
As the ball falls from a height to the floor, the gravitational potential energy stored in the ball will be converted to kinetic energy. The gravitational potential energy of the ball can be calculated using:gpe = mgh : m =0.8kg, g=9.8N/kg, h=5m so gpe = 0.8 x 9.8 x 5 = 39.2JKinetic energy can be calculated using: eK = 0.5 x m x v^2 . We can rearrange this to get v = SqRt ((2 x eK )/ m).eK = 39.2J, m=0.8kg, so v = SqRt ((2 x 39.2)/0.8) = 9.899 m/s. To 2sf this is 9.9m/s
Answer:
As the ball falls from a height to the floor, the gravitational potential energy stored in the ball will be converted to kinetic energy. The gravitational potential energy of the ball can be calculated using:gpe = mgh : m =0.8kg, g=9.8N/kg, h=5m so gpe = 0.8 x 9.8 x 5 = 39.2JKinetic energy can be calculated using: eK = 0.5 x m x v^2 . We can rearrange this to get v = SqRt ((2 x eK )/ m).eK = 39.2J, m=0.8kg, so v = SqRt ((2 x 39.2)/0.8) = 9.899 m/s. To 2sf this is 9.9m/s
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