During warm ups, a tennis player hits a ball which is 2.0 m above the ground. Th
ID: 1275362 • Letter: D
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 20.0m/s at an angle 5.2 degrees 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.
(1) At what time, after being struck by the racket, does the ball either pass over the net or else hit the net?
The answer ........... (seconds)
...................
(2) By how much does the ball clear the net? If the ball does not clear the net, but instead hits the net, enter this height as a negative number.
The answer: h=.........(m)
..................
(3) At what time, after being struck by the racket, does the ball reach its maximum altitude?
The answer is ...........(seconds)
.................
(4) What is that maximum altitude?
The answer is .......... (m)
..................
(5) 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.
The answer is .......... (seconds)
.................
(6) At what horizontal distance from the net does the ball finally hit the ground?
The answer is ............. (m)
.................
...(PLEASE, SHOW WORK IF POSSIBLE SO I GET HOW TO SOLVE IT).. THANK YOU.
Explanation / Answer
1)vx=20Cos5.2=19.92m/s
time=D/vx=7/19.92=0.35s
2)By equation of trajectory,
h=7Tan5.2-9.81*7^2/(2*20^2*Cos^2(5.2))=0.031m
But this is the height above the point of release. Height above net=1.031m
3)t=vy/g=20Sin(5.2)/9.81=0.185s
4)h=vy^2/2g=(20Sin (5.2))^2/(2*9.8)=0.17m
5)Using second equation of motion,
-2=20Sin5.2 *t-4.9*t^2
t=0.85s
6)R=vx*t=19.92*0.85=16.93m
Distance from net=16.93-7=9.83m
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