# Homework Help: Projectile Motion find the highest angle

1. Sep 6, 2014

### kukumaluboy

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

http://postimg.org/image/7yfw1zpq1/ [Broken]

http://postimg.org/image/7yfw1zpq1/ [Broken] <-Question

3. The attempt at a solution
I dont really understand the question but ill try.

I need to first find the highest angle at which the projectile will clear the peak.
At Peak:

Sy = 0.5(u+Vpeak)t , vpeak = 0
1800 = 125tsinx
tsinx = 14.4
t = 14.4/sinx -equation1

Sx = 0.5+(Ux+Vpeakx)t , Ux=Vx i guess?
2500 = 0.5(250cosx)t
2500 = 125tcosx
tcosx = 20
t = 20/cosx
Substitute eq 2 into eq 1:
20/cosx = 14.4/sinx
cosx = 1.39sinx

Then i dunno how to solve

Last edited by a moderator: May 6, 2017
2. Sep 6, 2014

### Orodruin

Staff Emeritus
You need to rethink a bit. It is not necessary that the vertical velocity is zero when passing the peak. Try finding the height as a function of the horizontal distance for an arbitrary angle.

3. Sep 6, 2014

### kukumaluboy

need more tips lol

4. Sep 6, 2014

### voko

This is a nice problem. I agree with Orodruin, that is probably the easiest method. What is your difficulty in following it?

5. Sep 6, 2014

### kukumaluboy

At top of parabola Vy = 0

Vy = Uy + at
0 = 250sinx + 9.81t
t = 25.5sinx

After that?

6. Sep 6, 2014

### Orodruin

Staff Emeritus
Yes, but the top of the parabola is not necessarily where the projectile passes the mountain top.

7. Sep 6, 2014

### SteamKing

Staff Emeritus
The top of the trajectory of the projectile can be reached before the peak is encountered, at the peak, or after the peak is encountered, so long as the altitude of the projectile clears the top of the peak.

8. Sep 6, 2014

### Orodruin

Staff Emeritus
How does this contradict what I just said?