# Projectile Motion, find max height, TWO POSSIBLE ANSWERS?

1. Aug 20, 2008

### iceman99

1. The problem statement, all variables and given/known data
A projectile is launched with an inital velocity of 30 m/s so that it covers a horizontal range of 45m. Determine the maximum height reached by the projectile. (There are two answers)

I know the answers are 2.96m and 43.0m. but I only know how to find the first. :(

2. Relevant equations
To be honest I have been working on this problem for a good 30 minutes. I found the 2.96m pretty easily, but its the 43m that has me stumped. I really have no idea what equations I could use to find the second maximum height of 43m.

3. The attempt at a solution
The only thing that I can see to solve the second part of the problem is that the projectile was thrown horizontally at 30 m/s from a building, landing 45m away. However when I put that theory to paper I came up with the height only being roughly 11.025m tall.
By this:
45m=30m/s times t
I found that t would equal 1.5s, after which i plugged that into the equation y=1/2gt^2 ending with y equaling 11.025m. But that is not right.

I tihnk the trick to this problem is in the wording of the problem but I just can't see it.

It would be greatly appreciated if I could have some help, this is for an AP physics summer assignment. I'm close to dropping this class because of how angry this is getting me. I would be grateful for any advice.

2. Aug 20, 2008

### rock.freak667

This is the entire question? If so, I don't think there can be two maximum heights for one parabolic motion.

3. Aug 20, 2008

### iceman99

Just tried another thing.
I set out to find the angle of the projectile when it landed setting my right triangle to:
Vx=30m/s
Vy=14.7m/s found by Vy=9.8m/s^2 x 1.5s Vy-14.7
Found the angle to be 26.10degrees

Making a new triangle with 26.10 as my theta and 45m as its adjacent length
Solving for the opposite length was found to be 22.045m darn...

I think something is just majorly wrong in my concepts for this to work, but oh well I was desprate for an answer.

4. Aug 20, 2008

### iceman99

Yeah that's the whole problem

5. Aug 20, 2008

### iceman99

Yeah I think that from the wording it means that it was launched from the ground at 30m/s which the maximum height was 2.96m. But because the question doesnt indicate that it wasnt launched from a building, it could be where the 43m comes from.

6. Aug 20, 2008

### Topher925

For every projectile launched at a fixed velocity and hitting a fixed target there are two angles in which it can do it. There will be one large on, and one small one. You found the small one, now you just need to find the other.

Pay careful to the equations that you use. The equation y = .5gt^2 you used, what does the "t^2" tell you about the solution of that equation?

Physics should not be hated, it should be enjoyed!

7. Aug 20, 2008

### iceman99

Um...I had no idea about that. How could I find the larger angle?

8. Aug 20, 2008

### Dick

Sure there are. If you launch it at 45 degrees it will reach a maximum range. If the given range is less than that then there is one angle greater than 45 degrees and one angle less than 45 degrees that will achieve that range.

9. Aug 20, 2008

### iceman99

OK, but how could I find the max height of the larger angle? Wouldn't the velocity need to be changed or am I wrong about that too?

10. Aug 20, 2008

### Dick

How did you find the smaller angle? You should be able to find the larger one the same way.

11. Aug 20, 2008

### iceman99

R=(Vo^2sin(2theta))/g i found theta to be 14.607degrees...

12. Aug 20, 2008

### iceman99

Oh I got it

13. Aug 20, 2008

### iceman99

I cant really believe I was that dumb. Sorry, thanks alot for the help

14. Aug 20, 2008

### Topher925

We have all done the same thing at one point.