A tennis ball is thrown horizontally from an elevation of 14.9 m above the groun

Question

A tennis ball is thrown horizontally from an elevation of 14.9 m above the ground with a speed of 18.8 m/s. (Let the initial position of the ball be the origin and down be the positive y-direction. Assume the ball is thrown in the positive x direction.) (a) Where is the ball after 1.50 s have elapsed? y = m x = m (b) If the ball is still in the air, how long before it hits the ground? (Enter 0, if the ball is on the ground.) s Where will it be with respect to the starting point once it lands? x = m

Explanation / Answer

a)

s = ut = 18.8*1.50 = 28.2 m

Vertical = 0.5*9.8*1.50^2 = 11.025 m

Thus From Ground = 14.9 - 11.025 = 3.875 m

b)

Let t be the time taken

Apply second equation of motion

s = ut + 0.5*g*t^2

14.9 = 0.5*9.8*t^2

Thus t = 1.74 sec

Horizontal Distance traveled = 18.8*1.74 = 32.71 m

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