What is the time taken by a ball thrown up to reach 98 m to come back to where it was thrown from.

Expert Answers

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Assume the ball is thrown in a direction perpendicular to the horizontal. The ball reaches a maximum height of 98 m. If the initial speed at which the ball was thrown upwards is U, `0 - U^2 = -2*9.8*98`

=> `U = sqrt(1920.8)`

The time taken by the ball to reach the highest point is `sqrt(1920.8) - 9.8*T = 0`

=> `T = sqrt(1920.8)/9.8 = 2*sqrt 5`

When the ball returns to the point it was thrown from, the speed of the ball is `sqrt(1920.8)` m/s. If the time taken for it to come down is `T'` , `-9.8*T'^2 = -sqrt(1920.8)`

=> `T' = 2*sqrt 5`

This gives the total time spent traveling as `4*sqrt 5` s.

The ball arrives at the point it was thrown from in `4*sqrt 5` seconds.

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