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?
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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