# Projectile motion of 2 dimensions

1. Jun 25, 2006

### Musicman

A man stands on the roof of a 17.0 m tall building and throws a rock with a velocity of magnitude 24.0 m/s at an angle of 33.0° above the horizontal. You can ignore air resistance.

(a) Calculate the maximum height above the roof reached by the rock.

(b) Calculate the magnitude of the velocity of the rock just before it strikes the ground.

(c) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

Ok I started out by doing 24sin(33) and 24cos(33). Then I did 17=0+23.99t+1/2(-9.8)t^2.
I got wrong answers for all 3 parts. I got 23.99 m for part A, for part B I got 44.21 m/s, and part C i got apprx 30 m. Can someone confirm my answers and not just show the answer alone, i need to know HOW to do it.

2. Jul 10, 2006

### Tom Mattson

Staff Emeritus
Hey folks,

Just a reminder of the PF Global Guidelines that we all agreed to, under Homework Help:

I've added the color for emphasis. Please don't post full solutions to problems. Helping is good. Doing the problem for the questioner is bad.

Thanks,

Tom

3. Jul 10, 2006

### Tom Mattson

Staff Emeritus
What is the y-component of the velocity when the rock reaches its maximum height?

You have functions that describe both the position and the velocity of the rock as functions of time. Because you know the coordinates of the rock just before it hits the ground, you should be able to find the time at which it hits the ground. Then you should be able to find the velocity at that time.

You're being asked about the x-coordinate when the rock hits the ground. You should know what time that is after completing part b.

Where did the 23.99 come from? You're supposed to have $v_{0y}$ in that spot, but $v_{0y}$ certainly doesn't equal 23.99 m/s.

Try to follow the hints I gave. If you're still stuck then post what you've got and we'll try to get you un-stuck.