# Projectile Question

1. Sep 24, 2005

### SS2006

im completely puzzled, dont nkow what to do, so frustratiing :(

heres the pic, ive done one like this beofre, but i had eth angle it was fired at, this time we dont

2. Sep 24, 2005

### Jameson

Separate the velocity vector into its two components.

$${V_{x0}}={V_0}\cos{\theta}}$$
$${V_{y0}}={V_0}\sin{\theta}}$$

You can then solve for the total time it will take to hit the ground by working with the y-component, and thus use the x-component to solve for the x displacement.

Last edited: Sep 24, 2005
3. Sep 24, 2005

### SS2006

lets say i want to to the v(x)
id get
v(x) = 5 cos ?
i dont know the degree, usually i have 5 cos 60 degrees, then id have a V(x), this time id ont

4. Sep 24, 2005

### Päällikkö

As it's fired horizontally, what's the angle?

5. Sep 24, 2005

### Jameson

Use one of your kinematic equations.... how about $$V_{yf}^2=V_{y0}^2+2gd$$. All of those variables, and gravity, are given to you. What can you solve for now?

6. Sep 24, 2005

### SS2006

180 degrees right, tried that and i got real close, just a 0.01 away, think i got it
thx

7. Sep 25, 2005

### SS2006

can i use 180 degrees?

8. Sep 25, 2005

### Päällikkö

I'd use 0 degrees. With 180 you have to hassle with directions (you get negative cosine).