Your opponent, in a desperate attempt to return your wicked serve, makes a high

Question

Your opponent, in a desperate attempt to return your wicked serve, makes a high lob of the tennis ball that causes it to land out of bounds on the asphalt course behind you. It then bounces so that it just barely clears a 3.90 m wall that is 2.20 m from where the ball hit the asphalt and that is separating your court from the next. In the adjacent court play is suddenly interrupted by the ball leaving your court and bouncing 2.51m up into the action. How far horizontally, in meters, does the ball travel in this second bounce? Hint: Use the trajectory of the first bounce to get the horizontal speed, and assume that it remains unchanged between bounces.

Explanation / Answer

let initial speed afte first bounce is v and angle above horizontal is theta.

maximum height achieved is 3.9 m

==>v^2*sin^2(theta)/(2*g)=3.9

==>v*sin(theta)=sqrt(3.9*2*g)=8.743 ....(1)

time taken for the ball to reach ground level again is given by 2*v*sin(theta)/g=1.7843 seconds

then horizontal distance travelled=horizontal veloicty*time

=v*cos(theta)*1.7843=2.2

==>v*cos(theta)=1.233 ...(2)

dividing equation 1 by equation 2:

tan(theta)=8.743/1.233=7.1

==>theta=81.973 degrees

then substituting the same in either equation 1 or 2, we get

v=8.83 m

as given in the problem, the horizontal velocity remains constant

hence for the next bounce, horizontal veloicty=v*cos(theta)=1.233 m/s

if for the second bounce, velocity is v1 and angle above horizontal is theta1,

then v1*cos(theta1)=1.233....(3)

maximum height=2.51

==>v1^2*sin^2(theta1)/(2*9.8)=2.51

==>v1*sin(theta1)=7.014 m/s....(4)

horizontal distance is given by

v1*cos(theta1)*2*v1*sin(theta1)/g

=1.233*2*7.014/9.8

=1.765 m

hence horizontally it will cover 1.765 m

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.