How Do You Solve a Kinematics Tennis Ball Problem Using d-v-a-t Formulas?

[dragonx47]

Homework Statement

) 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 x2Identify 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.

Homework Equations

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

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.

 

[RUber]

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.