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A ball is thrown straight upward and returns to the thrower's hand after 3.05 s in the air. A second ball thrown at an angle of 28.0° with the horizontal reaches the same maximum height as the first ball. At what speed was the first ball thrown?

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akande212

Answer:

U = 17.15m/s.

Explanation:

The ball rises up a height h. After reaching this height the ball falls back down this same height. The time taken to rise the height h is also equal to the time taken to fall the same height.

So the total time taken for the flight 3.50s is equal to 2× the time taken to rise to height h, t.

See the attachment below for the complete solution.

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