# Homework Help: Projectile ball on curved runway (conservative forces)

1. Sep 23, 2007

### thatgirlyouknow

1. The problem statement, all variables and given/known data

A particle, starting from point A in the drawing (the height at A is 3.00 m), is projected down the curved runway. Upon leaving the runway at point B, the particle is traveling straight upward and reaches a height of 7.53 above the floor before falling back down. Ignoring friction and air resistance, find the speed of the particle at point A.

Ok the picture looks like this (I apologize for the slightly craptastic drawing):

o <--ball (at 7.53 m)

^
| A (3.00 m)
| /
*B* /
L___/

2. Relevant equations

Conservation of energy
PE = mgh
KE = (1/2)mv^2

3. The attempt at a solution

I started off with the basic equations of kinematics (like finding displacement based on v0 and vf) only to be told "You do not need to know the speed or vertical position of the particle at point B in order to solve this problem." Therefore, I am clueless where to go from here.

2. Sep 23, 2007

### learningphysics

Use conservation of mechanical energy.

3. Sep 23, 2007

### Proggle

You almost have it solved.

PE(A) + KE(A) = PE(B) + KE(B)

Substitute the equations you mentioned with the appropriate values and solve for v at A.

Last edited: Sep 23, 2007
4. Sep 23, 2007

### thatgirlyouknow

Ef = E0. So either mgh0=mghf or .5mv0^2=.5mvf^2
Mass is the same in both, so
.5v0^2=.5vf^2 or
gh0=ghf

.5mvf^2 + mghf = .5mv0^2 + mgh0
removing the constant m
.5vf^2 + ghf = .5v0^2 + gh0
.5*0 + -9.8*7.53 = .5*v0^2 + -9.8*3
-44.394 = .5v0^2
88.788 = v0^2
v = 9.423

Thank you!