A ball is thrown horizontally with a speed of Vo = 20.0 m/s from the top of a bu

Question

A ball is thrown horizontally with a speed of Vo = 20.0 m/s from the top of a building of height h = 42.1 meters. When the ball hits the ground, it is traveling with a speed of 35.0 m/s. What is angle below the horizontal made by the velocity vector of the ball when it hits the ground? What horizontal distance does the ball travel before it hits the ground?

Explanation / Answer

We generally do not answer the questions of those who are having low rating. You are having a rating of "73%". Please rate all answers and improve your rating. Even if you find the answers are incorrect, rate them as not helpful. But do rate them for the good of both of us. Thanks!! Coming to the problem Horizontal velocity = 20 m/s as there is no acceleration in horizontal direction. Vertical velocity = sqrt((35^2)-(20^2)) m/s Tan(teta) = Vertical velocity/Horizontal velocity = sqrt((35^2)-(20^2))/20 =>teta = 55.1501 degrees Initial vertical velocity = 0 m/s = u Final vertical velocity = sqrt((35^2)-(20^2)) m/s = v acceleration = g = 9.81 m/s^2 = a v = u+at =>t = v/a = sqrt((35^2)-(20^2))/9.81 = 2.927911645 So horizontal distance traveled = Horizontal velocity*t = 2.927911645*20 = 58.56 m

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