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A lead ball is dropped onto a diving board from a height of 2 meters. The ball rebounds to a height of 1.5 meters. If the ball is in contact with the diving board for 0.1 seconds, what is the average acceleration of the ball during contact?

1) 2 m/s²
2) 3 m/s²
3) 4 m/s²
4) 5 m/s²

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Final answer:

The average acceleration of the ball during contact with the diving board is 5 m/s².

Step-by-step explanation:

To find the average acceleration of the ball during contact with the diving board, we can use the formula:

acceleration = (change in velocity) / (time)

Since the ball is dropped from a height of 2 meters and rebounds to a height of 1.5 meters, the change in velocity is the difference in these heights. The change in velocity is 2 - 1.5 = 0.5 meters per second. The time of contact is given as 0.1 seconds. Plugging these values into the formula, we get acceleration = 0.5 / 0.1 = 5 m/s². This computation highlights the acceleration experienced by the ball during the interaction, crucial in analyzing forces and impacts in physical scenarios like collisions or rebounds.

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