Cannon Projectile Motion Problem

[vanquish]

[SOLVED] Projectile Motion

Homework Statement

You are shooting a cannon from a height of 11.23m off the ground. The target is 20.36m away. The angle of projection is 30 degrees. What does the initial speed need to be in order to hit the target?

Homework Equations

When I look at this problem I just cannot solve it. I personally don't think I have enough information but it has really been bugging me.

The Attempt at a Solution

I know I need to figure out one of the velocities, either for the x or the y axes in order to solve for the initial velocity. The only information I have is:
$$\Delta$$y=11.23
a=-10

$$\Delta$$x=20.36
a=0

I tried using the range formula before remembering that it can only be used when the initial and final velocities are the same.

 

[rocomath]

You have everything you need.

$$x=(v_0\cos\theta)t$$
$$y=(v_0\sin\theta)t-\frac 1 2 gt^2$$

Solve for t in the x and plug it into y, then solve for v initial.

 

[vanquish]

Thanks very much, that makes perfect sense. I'll keep a lookout for problems like these in the future

 

[rocomath]

Thanks very much, that makes perfect sense. I'll keep a lookout for problems like these in the future

Welcome. You may also want to write those formulas down in your book for future reference so you don't have to go through the annoying algebra everytime.

 

[Yea]

You have everything you need.

$$x=(v_0\cos\theta)t$$
$$y=(v_0\sin\theta)t-\frac 1 2 gt^2$$

Solve for t in the x and plug it into y, then solve for v initial.

I just tried it and the answers does not make sense.

 

[rocomath]

I just tried it and the answers does not make sense.

What did you get?

 

[Yea]

I got vo= 72

 

[rocomath]

I got vo= 72

Final equation should be ...

$$v_0=\frac{x}{\cos\theta}\sqrt{\frac{g}{2(x\tan\theta-y)}}$$

 

[vanquish]

I got the same problem... maybe our algebra is the problem

 

[rocomath]

$$y=v_0\sin\theta\left(\frac{x}{v_0\cos\theta}\right)-\frac{1}{2}g\left(\frac{x}{v_0\cos\theta}\right)^2$$

$$y=v_0\tan\theta-\frac{gx^2}{2(v_0\cos\theta)^2}$$

 

[vanquish]

Final equation should be ...

$$v_0=\frac{20.36}{\cos\(30)}\sqrt{\frac{-10}{2(20.36\tan\(30)-11.23)}}$$

This is the equation with all the numbers plugged in, and its still not turning out right

(should be cos 30 and tan 30, but i couldn't get it to work)

 

[Yea]

what is up with this problem, why doesn't the answer make any sense?

 

[vanquish]

The algebra looks all right, so I don't understand how we could be getting such a wrong answer. I have a feeling it's a simple little mistake

 

[Yea]

i have a feeling that this problem cannot be that complicated.

 

[rocomath]

The algebra looks all right, so I don't understand how we could be getting such a wrong answer. I have a feeling it's a simple little mistake

What's the actual answer?

 

[vanquish]

we forgot the negative on the delta y...

 

[Yea]

ahhhh that's the key

 

[Yea]

we forgot the negative on the delta y...

ahh that's the key

 

[rocomath]

we forgot the negative on the delta y...

You're the man or woman! :-]