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Kinematics Tennis Ball Problem

  • Thread starter dragonx47
  • Start date
1. The problem statement, all variables and given/known data
) You are given the task of shooting a tennis ball from ground level through 2 hoops. The two hoops’ centers and the launch site are located in the same vertical plane, and the hoops are oriented perpendicular to the ball’s proposed trajectory and also in a vertical plane.

The first hoop has a height y1 and is located at a horizontal distance from the x1 launch site (which is located at x0, y0). The second hoop is located at x2, y2.

a) Use the d-v-a-t formulas to eliminate time and solve for the y-position as function of the x-position. In particular, show that

y = a x + b x2

Identify the quantities a and b in terms of launch speed v0 and launch angle q0, and the gravitational field strength g (which will later take on the value of 10 m/s2).

b) Solve for a and b in terms of x1, y1, x2, and y2.

c) For the case of y1 = 4.0 m, x1 = 2.0 m and y2 = 3.0 m, x2 = 4.0 m, determine the values of v0 and q0.

2. Relevant equations
y = Vy*t + .5(g)(t)^2
x = Vx*t
Vx = Vocos(theta)
Vy = Vosin(theta)

3. The attempt at a solution
I've solved the a) part of the question and gotten that the coefficient a = tan(theta) while the coefficient b = g/(2*(Vocos(theta))^2). I'm not sure how to put these two equations in terms of the position coordinates x1, x2, y2, and y1.


Homework Helper
Are you assuming the ball's trajectory will pass through the centers of the hoops? If so, then you have 3 points, (x0,y0), (x1,y1), and (x2,y2), to solve what appears to be a quadratic form.
Also, since simple ballistic motion is symmetric, you should see that you want the max y value to occur at (x1+x2)/2, so, you should be able to back that out into some requirement for q0 and v0.

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