# Projectile calculation help

1. Oct 5, 2004

### Format

Need help with this one:

A Projectile is launched at 35.0 degrees above the horizontal with an intial velocity of 120m/s. What is the projectiles speed 3.00 seconds later?

Ive got the x and y compontents...wut now?

2. Oct 5, 2004

### faust9

This is what you have:
$$\vec v=v_xi+v_yj$$

Think of the above like a right triangle with the speed as the hypotenuse. How would you go about finding the length if this were a trig or algebra class?

3. Oct 5, 2004

### UrbanXrisis

The projectial's horizontal speed would be the x component
Find the vertial speed by using

a=-9.8m/s^2
t=3s
SpeedInital=y-component
SpeedFinal=solve for this

4. Oct 5, 2004

### Format

well i tried using Vf = Vi + at ...which has all i need but it doesnt give the correct answer (106m/s)

5. Oct 5, 2004

### Format

ohh, nevermind i get it...its the resultant of the 2 components thx

6. Oct 5, 2004

### faust9

It wont give you the correct answer. The speed of the projectile is the magnitude of the velocity vector. Velocity vector has two components--velocity in the x direction and velocity in the y direction. You said you found the x and y components already so you have your vector components.

7. Oct 5, 2004

### faust9

Good. You figured it out before I finished typing.

8. Oct 5, 2004

### Format

lol yea thx! One other thing concerning projectiles...how do you solve for maximum height with only an initial velolcity at a certain angle?

9. Oct 5, 2004

### Pyrrhus

At Max Height, Vy will be 0, you know the vector is tangent to the parabole, if you know your calculus, you know the vector will be horizontal at max height.

10. Oct 5, 2004

### Format

hmm...still not gettin it

11. Oct 5, 2004

### Pyrrhus

12. Oct 5, 2004

### Format

Y Components:
Vi = 7.89
Vf = 0
a = -9.8
t = 1.62
d = ?

d = Vit + .5(a)t^2
0 = 7.89t + .5(-9.8)t^2
t = 1.62

d = 7.89(1.62) + .5(-9.8)(1.62)^2

better?

13. Oct 5, 2004

### Format

lol nevermind im dumb...i got it