# Finding the vertical component of velocity

1. Oct 9, 2011

### shyguy79

The question simply states that a cannonball is fired over a wall 40m away and passes directly over the wall at a height of 15m. Find the vertical component of velocity?

I've been searching the net and apparently when V sub y =0 but I really don't understand this? Any help would be great!

2. Oct 9, 2011

### Villyer

If you don't know where to start, its usually a good idea to write out what you know and the equations that relate to the problem.

For projectiles problems, separate the givens into two columns, the x component and the y component. This helps you keep them organized and separate, since only time is a common element of the two.

3. Oct 9, 2011

### shyguy79

Looking at the equations I've got to play with...

T=sqrt(2*Sy/a) where Sy is the vertical displacement and a=acceleration 9.81ms^-2 so putting my figures in

T for the ball to reach the highest point would be: sqrt(2*15/9.81) = 1.75sec

4. Oct 9, 2011

### shyguy79

Using T= 1.75 and rearranging the equation.. Vy=Uy+aT into Uy=Vy-aT where Uy is the initial velocity, Vy is the velocity at the top, a is the acceleration due to gravity.

Uy=0+9.81*1.75 = 17.2 ms^-1 (1 dp)

Incidentally if I use v^2=u^2+2aSy and rearrange for u=sqrt(v^2-2aSy) = 17.2 so both answers are the same does this mean I'm on the right track????

But I'm not sure if this is correct...

Last edited: Oct 9, 2011
5. Oct 10, 2011

### shyguy79

No help here then... Thanks anyways