Question

A tennis player hits a ball from a height of 1.5 meters. The ball leaves his racket with speed 6.0 m/s at an angle of 10◦. If the net is 5.0 meters away and 1.0 meter high, will the ball pass above the net?

Answer #1

Given that,

h = 1.5 m

vi = 6 m/s @ 10 degrees

vx = 6cos10 = 5.90 m/s

vy = 6sin10 = 1.041 m/s

Since net is 5 m away. so ball has to travell horizontal distance more than 5 m to pass above the net.

From kinematic equation,

h = ut + (1/2)at^2

Consider upward direction +ve.

-1.5 = 1.041*t - (1/2)*9.8*t^2

4.9t^2 - 1.041t - 1.5 = 0

By solving quadratic equation,

t = 0.6696 s

Horizontal distance covered by the ball,

x = ux*t

x = 5.90*0.6696

x = 3.95 m

Since Horizontal distance is less than 5 m so ball will not pass above the net

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