Recent content by jamescb82

Thanks for the help!

Yes you're right! Thank you for catching me on that so t should be 1.35s which brings the final velocity to 10.5 m/s

Xf=range (distance from net = 10.0m) Yf=displacement above initial launch point (3.05m - 2.00m = 1.05m) VA=initial velocity

So I attempted to solve for the final velocity and thought about the horizontal and vertical components of this problem. In the end I took these equations and used substitution to solve for time (1.83s) and used that time to solve for velocity initial (7.73m/s). I don't have the answers to these...

If I numbered these equations 1 - 4, top to bottom, I'm guessing number two and four?

Homework Statement So I am given the height of a basketball player (2.00m tall) and the height of a basketball net (3.05m tall) and the distance at which the player stands from the net (10.0m away). The player shoots the ball at an angle of 45.0 degrees. What must the initial velocity be to...