# Homework Help: Help with projectile motion, finding x

1. Oct 2, 2014

### navm1

1. The problem statement, all variables and given/known data
A balloon is attached to a 3m high pole. On the ground is a mortar that fires at a velocity of 7m/s at an angle of 55degrees. How far away does the mortar have to be to hit the balloon?

2. Relevant equations
y=tan(theta)x-g/(2v0cos2(theta)x2

3. The attempt at a solution
I rearranged to make a quadratic equation from the equation above and made

g/(2v0cos2(theta)x2 - tan(theta)x+y=0

-9.81m/s2/(2*(7m/s)2(cos2(55deg)x2 - tan(55)x+3m

and ended up with 1.58m and -6.34m

I understand that there will be two solutions because the projectile can hit the balloon on the way up and the way down but i dont think these answers are correct.

2. Oct 2, 2014

### vela

Staff Emeritus
You seem to have a sign error. However, with the given numbers, I find there's no solution.

3. Oct 2, 2014

### navm1

in my lecture the correct answer was something like 1.9 or something and i forget the other one but when i plugged my answer 1.58 back into y=tan(theta)x-g/(2vcos^2(theta)x^2 it comes out as the right y value - 3m.

is there a chance im using the completely wrong equation for this?

4. Oct 2, 2014

### vela

Staff Emeritus
You're using the right equation. Here's a plot of the trajectory with the given numbers. The projectile never reaches a height of 3 m.

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5. Oct 2, 2014

### vela

Staff Emeritus
When I plug in x=1.58 m, I get y=1.50 m.

6. Oct 2, 2014

### navm1

I feel a little lost then. so for a projectile firing at 7m/s at 55 degrees trying to hit a balloon 3m in the air, it will never actually hit it?

7. Oct 2, 2014

### vela

Staff Emeritus
Right. It's not going fast enough to reach a height of 3 m when it's aimed at 55 degrees off the horizontal. In fact, even if you shot it straight up, it wouldn't reach that height.

8. Oct 2, 2014

### navm1

im following now. max height is 1.68m if i worked it out correctly so perhaps my notes are wrong. I'll compare notes with a friend and post again. thanks for your help so far

9. Oct 2, 2014

### vela

Staff Emeritus
I'm not sure how you got that number. There's something weird going on with your calculations.

10. Oct 2, 2014

### navm1

for max height i used

v0^2sin^2(theta)/2g

72*sin2(55) / 2 * 9.81 = 1.676

11. Oct 2, 2014

### vela

Staff Emeritus
Ah, okay, I thought you were referring to shooting the projectile straight up.