I think a problem might be that you are finding t by finding when the final velocity is equal to zero. This occurs at the maximum height, which does not represent the time it takes to fall to the ground. I would try finding an expression for t from your x direction information.
Hage567, that makes sense!!!! Absolutely, the velocity in the y-direction is zero at two different points, at y-max and when it hits the ground. If I solve is using the x-direction information I can find the time of flight, t=4.54 s. Now the next step is to solve for the initial velocity, using the x-direction equation I find v0=19.44 m/s^2 but why can't I use the y-direction equation vf=v+at ? If I use this I get v0 = 104.45!! Is it for the same reason? That this equation can not be used reliably when vf=0 because it occurs at 2 different points?
You have to be careful when saying vf is zero at two points. vy is instantaneously zero at the maximum height, but at the moment when it hits the ground, the velocity is not zero. So you can't use vf=v+at in that manner. How did you get the time of flight from your x direction, if you don't have v yet? I'm not sure what you did there. Basically, you should end up with two equations and two unknowns (v and t), which you can easily solve.
I used the x-direction equation: change in x= V0*cos(25)*t. First I solved it for V0 and then I plugged it into the y-direction equation: change in y=V0*sin(25)*t-1/2gt^2. This is what you are suggesting right? Perhaps I was not listening in class, but is it incorrect to say that the Vf is zero when it hits the ground right? I know its not zero at that instant. You cant base any calculations on the fact that it is zero an instant after it hits the ground? In that case, vy is only zero when it is at its max height, when it changes direction. So when can you say vf is zero? (thanks by the way, you are being very helpful, I really appreciate it)
“I used the x-direction equation: change in x= V0*cos(25)*t. First I solved it for V0 and then I plugged it into the y-direction equation: change in y=V0*sin(25)*t-1/2gt^2.”
Yes I think this looks OK (I don’t have my calculations with me and I don’t have time to work it out again, so I have to go by memory).
I guess the reason you can’t use Vy=0 for when it hits the ground is because for v to become zero, the ball (or whatever) would have to undergo a deceleration (due to the force of hitting the ground). This deceleration is not “g”, so you would have to be able to quantify it to be able use it. So take Vy as the velocity of the object at the instant it the ground. Does that make sense?
Yes, that makes sense. When the object is moving "downward" g is an acceleration, so it does make sense that its velocity is increasing (in contrast when g acts like a deceleration when the object is moving upward, right?). So since we take vy as the velocity at the instant it hits the ground, vy is never zero at that point. I want to make sure I never make that mistake again, using vy=0 when it hits the ground. Thanks.