Projectile motion: mortar fired at balloon 3m high

[navm1]

Homework Statement

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?

Homework Equations

y=tan(theta)x-g/(2v₀cos²(theta)x²[/B]

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 don't think these answers are correct.

 

[vela]

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

 

[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 I am using the completely wrong equation for this?

 

[vela]

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.

 

[vela]

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.

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

 

[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?

 

[vela]

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.

 

[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

 

[vela]

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

 

[navm1]

for max height i used

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

7²*sin²(55) / 2 * 9.81 = 1.676

 

[vela]

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