A tennis ball is thrown horizontally from an elevation of 14.2 m above the groun

Question

A tennis ball is thrown horizontally from an elevation of 14.2 m above the ground with a speed of 21.7 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 = 32.6 Incorrect: Your answer is incorrect. m x = 11.0 Incorrect: Your answer is incorrect. 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.) 1.7 Incorrect: Your answer is incorrect. s

Explanation / Answer

a)
so horizontal:

x = vt =21.7m/s * 1.50s =32.55 m

vertical:
since the ball is thrown horizontally, the initial vertical velocity is 0.
y =1/2*at^2=1/2*9.8*(1.50)^2= -11.036 m
y =14.2 -11.036 = 3.16 m above the ground

b)
y=4.9t^2
14.2 = 4.9t^2
t=sqrt of (14.2 / 4.9) =1.70s
It will take 0.20 more seconds than from the place in part a

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