A ball is thrown horizontally from the top of a 95 m building and lands 150 m fr

Question

A ball is thrown horizontally from the top of a 95 m building and lands 150 m from the base of the building. Ignore air resistance, and use a coordinate system whose origin is at the top of the building, with positive y upwards and positive x in the direction of the throw.

Part (b) What must have been the initial horizontal component of the velocity in m/s?

Part (c) What is the vertical component of the velocity just before the ball hits the ground in m/s?

Part (d) What is the magnitude of the velocity of the ball just before it hits the ground in m/s?

Explanation / Answer

a) H=95

H= 0.5 x 9.8 x t^2

=> t =4.403s

Thus, it covers v x t horizontal distance .

150 m = Vx x 4.403

=> Vx= 34.066 m/s   

Thus, the horizontal component of the velocity is 34.066m/s

c) vertical component of velocity just before touching the groung

Vy= sqrt(2 x g x H) = - ( 2 x 9.8 x 95) = -43.15m/s

d) Magnitude of velocity = sqrt(Vx^2 + Vy^2 ) = 54.977 m/s

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