# Homework Help: Projectile Motion

1. Jan 30, 2008

### jesuslovesu

1. The problem statement, all variables and given/known data
http://img524.imageshack.us/img524/6899/tempnb3.th.png [Broken]

2. Relevant equations
y = y0 - 1/2gt^2 + voy*t

3. The attempt at a solution
Well there are two methods that I used for solving this, but I think only one is correct:

First, the correct method:
1) Find the time at the top of the parabola and use the distance from the hoop to determine the time.
12 = 5 - 16t^2 + v0*t*(5/13)
x = v0*(12/13)*t = 24
t = .433s

v0 is 60 ft/s

2) Find the time at y = 3 (to hit the second slope)
0 = 5-3 - 16t^2 + vo(5/13)*t
t = 1.524s

3) Find h
x = vo(12/13)*t = h + 24 --> h = 60.4 ft

I think everything I did is correct, however, my TA in lecture today said that the v0 should be between 50 ft/s - 60 ft/s and h should be between 40 ft and 47 ft. However, he also said that he didn't quite know how to get the correct solution.... (he was looking at the solutions manual)

I get 60ft/s but my h is 60.4 ft however, if I use the 2nd method I get within the range, but isn't my method correct and the second method wrong?

method 2:
12 = 5 - 16t^2 + v0*t*(5/13)
v = vo(5/13) - gt
v = 0 since it's at the top of it's path
v0 = gt * 13/5
v = 55 ft/s ...
t = .661 s
But isn't that wrong because if I plug in .661 s into x = vo(12/13)*t I get ~33ft, whereas it should be 24 ft at the top of the path..

Can anyone advise me on which method is correct?

Last edited by a moderator: May 3, 2017