Kinematics-acceleration in two dimensions

[PEZenfuego]

Homework Statement

A tennis player standing 8.0 meters from the net hits a ball 1.5 meters above the ground toward her opponent. The ball leaves her racquet with a speed of 25.0m/s at an angle of 14.0 degrees above the horizontal. The net is 1.0 meters high. The baseline is 12 meters back from the net; a ball striking the ground beyond the baseline is out of play

Homework Equations

x=V.costheta(t)
y=V.sintheta(t)-0.5g(t^2)

The Attempt at a Solution

I attempted to solve for t using the quadratic equation in conjunction with the above formula for y. I ended up with two solutions, neither was extraneous from what I could tell. I plugged both in for x and ended up with two answers. One was slightly more than 8 (meaning that the ball landed in), the other was a little over 21 (meaning that the ball landed out). I am lost, which is it and how do I know?

 

[gneill]

You need to include the initial height of the ball (where its launched) when you work with the vertical component of the position.

 

[PEZenfuego]

You need to include the initial height of the ball (where its launched) when you work with the vertical component of the position.

Right, and I did. I got the equation deltaY=V.sintheta(t)-0.5gt^2

Rearranged and solve the quadratic equation. I ended up with two answers. One was 0.89 and the other was 0.3437. How am I supposed to determine which is correct and which is not?

 

[gneill]

Right, and I did. I got the equation deltaY=V.sintheta(t)-0.5gt^2

Rearranged and solve the quadratic equation. I ended up with two answers. One was 0.89 and the other was 0.3437. How am I supposed to determine which is correct and which is not?

I'll bet you've set your Δy to a positive value, right But doesn't it FALL from its initial position to hit the ground?

 

[PEZenfuego]

:facepalm: I appreciate the help. I remember running into this same problem in high school. Maybe after learning from the same mistake twice, I won't make it again. Fingers crossed.