During warm ups, a tennis player hits a ball which is 2.0 m above the ground. Th

Question

During warm ups, a tennis player hits a ball which is 2.0 m above the ground. The ball leaves his racquet with a speed of 16.0 m/s at an angle 4.6 above the horizontal. The horizontal distance to the net is 7.0 m, and the net is 1.0 m high. Assume that the path of the ball, as observed from a bird's eye view, is parallel to the long lines of the tennis court.

A- What is that maximum altitude?

B-At what time, after being struck by the racket, does the ball finally hit the ground? Note: if the ball would have hit the net, assume that the net has a hole in it at that location so that the ball would pass through the net with no interruption in its flight path.

C- At what horizontal distance from the net does the ball finally hit the ground?

Explanation / Answer

(A)

From kinematic equation,

Maximum altitude,

h = u^2 / 2g

h = (16*sin4.6)^2 / 2*9.8

h = 0.084 m

Hmax = 2 + 0.084

Hmax = 2.084 m

(B)

from the equation,

h = uy*t - (1/2)gt^2

-2 = 16*sin4.6*t - (1/2)*9..8*t^2

4.9t^2 - 1.28t - 2 = 0

b solving the qudratic equation,

t = 0.783 s

After 0.783 s, ball finally hits the ground

(C)

R = vx*t

R = 16*cos4.6*0.783 = 12.48 m

horizontal distance from the net ,

d = 12.48 - 7

d = 5.48 m

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