A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.30 m/s. If the roof is pitched at 22.0° below the horizon and the roof edge is 2.10 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground.

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Asked Oct 11, 2019
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A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.30 m/s. If the roof is pitched at 22.0° below the horizon and the roof edge is 2.10 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground.

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Expert Answer

Step 1

Given information:

Velocity of the ball before leaving the roof (u) = 4.30 m/s

Angle of the roof (θ) = 22.00

Height of the roof from the ground (h) = 2.10 m

Step 2

Time the ball spent in are:

Consider 2nd...

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h (usin0)t 9t2 where "usin0" is the vertical component of the velocity, "g" is the acceleration due to gravity Substitute the corresponding values, we get: 1.61t4.9t2 2.1 4.9t21.61t - 2.1 = 0 We get "t" as t 0.51 s (or) -0.83 s ve" value of time is not considered Time spent in air by the ball is 0.51 s

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