During a tennis match, a player serves the ball at 26.4 m/s,with the center of t

Question

During a tennis match, a player serves the ball at 26.4 m/s,with the center of the ball leaving the racquet horizontally 2.36 mabove the court surface. The net is 12.0 m away and 0.900 m high.When the ball reaches the net, (a) what is the distance between thecenter of the ball and the top of the net? (b) Suppose that,instead, the ball is served as before but now it leaves the racquetat 5.00° below the horizontal. When the ball reaches the net,what now is the distance between the center of the ball and the topof the net? Enter a positive number if the ball clears the net. Ifthe ball does not clear the net, your answer should be a negativenumber.

Explanation / Answer

in the first case horizontal velocity of the ball=26.4 m/s vertical velocity=0 vertical accleration=g=9.8 time taken to travel 12 m=12/26.4 =0.45 s in 0.45 s let the ball travelled h m downwards h=u*t+(1/2)*g*t2 h=(1/2)*9.8*0.45*0.45 h=1.012 m the ball is 2.36-1.012 =1.348 m the ground ir it is1.348-0.9=0.448 m above the net ie the distance between the net and center of the ball= +0.448 m(ans a) now horizontal velocity=26.4*cos 5 vertical velocity=26.4 sin5 time taken to travel 12 m=12/(26.4*cos5)=0.46 s in 0.46 s the ball travelled h m downwards h=26.4*sin5*0.46+(1/2)*9.8*0.46*0.46 h=2.096 m the height of the ball above the ground=2.36-2.096 =0.263 m the ball is 0.9-0.263=0.637 m below the net distance between the center of the ball and top of the net= -0.637m (ans b)

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