# Homework Help: Projectile Motion

1. Jun 20, 2009

### theuniverse

1. The problem statement, all variables and given/known data
A bullet is fired from a cliff 36m above the ground. If the bullet is fired at an angle of 25 above the horizontal, and has a muzzle velocity of 80m/s, what is the velocity of the bullet as it hits the ground?

d = 36m
vi = 80m/s
angle = 25 degrees

2. Relevant equations

3. The attempt at a solution
Horizontal: 80 cos 25 = 72.5

Vertical: muzzle: 80 sin 25 = 33.8

0^2 = 33.8^2+2ad <-- Top of the projectile motion
0= 1142.44+2(-9.8)d
d= 1142.44/19.6
d= 58.28

height = 58.28+36 = 94.28
Vf^2= vi^2+2ad <-- Top of the parabola = 0m/s
vf^2= 2(9.8)(94.28)
Vf^2= 1847.88
Vf= 42.99
R^2 = 42.99^2+72.5^2 = 7105.25
R = 84.3

I was just wondering if I did it right...

Thank you very much,

2. Jun 20, 2009

### Staff: Mentor

Looks good to me, but you could have calculated the change in vertical speed in one step using the same formula (Vf^2 = Vi^2+2ad). Let a = -9.8 m/s^2 and d = -36 m.

(You could also have used conservation of energy.)