# Projectile motion

1. Oct 2, 2008

### kitle545

1. The problem statement, all variables and given/known data

A projectile is fired at an angle of 60 degrees above the horizontal with an initial speed of 300 m/s. Calculate A) the horizontal distance traveled and B) the vertical height attained in the first 6 s.

2. Relevant equations

3. The attempt at a solution
I did 300cos60=150 so Vi(x)=150 and 300sin60=259.8 so Vi(y)=259.8 but after that it seems like i don't have enough information to continue.

2. Oct 2, 2008

### Rake-MC

you need to use constant acceleration formulae

we know:
Ux = 150ms-1
Uy = 259.8ms-1
a = -9.81ms-2

First you can calculate how long it's in the air for, we know that its trajectory is symmetrical about the peak, and at the peak it's y velocity is 0ms-1.

$$v = u + at$$

$$t = \frac{v-u}{a}$$

= $$\frac{0 - 259.8}{-9.81} = 26.48s$$

For it to hit the ground again, it must have been in the air twice that time = 53s.

now you know for the x plane, a, t and u
find s (displacement)

$$s = ut + \frac{1}{2}at^2$$

hint: half of that equation dissapears when you sub in the values.

You can do the rest yourself!

Last edited: Oct 2, 2008