# Motions on 2 dimensions

1. Sep 30, 2008

### -EquinoX-

1. The problem statement, all variables and given/known data
A playground is on the flat roof of a city school, 6.00m above the street below. The vertical wall of the building is 7.00 m high forming a 1m high railing around the playground. A ball has fallen to the street below and a passerby returns it by launching it at an angle of 53 degrees above the horizontal at a point 24 m from the base of the bulding wall. The ball takes 2.20 to reach a point vertically above the wall.

Find the vertical distance by which the ball clears the wall.

2. Relevant equations

3. The attempt at a solution

I can get the initial velocity first by plugging in the formula:

24 = Vi*cos(53)*(2.2) then

finding for Vi I get 18.1. Now how do I find the height the ball clears the wall?

Last edited: Sep 30, 2008
2. Oct 1, 2008

### Kurdt

Staff Emeritus
You need an equation that has the distance the acceleration and the initial and final velocities in the y direction. You know at the max height the velocity in the y direction will be zero. You know the initial velocity and the acceleration and thus you can work out the height.