# Homework Help: How to find the initial velocity given only angle and distance traveled

1. Sep 19, 2010

### ://Justice

1. The problem statement, all variables and given/known data
An object is shot (from a cannon) at an angle of 33 degrees and landed 85 m away. Calculate the magnitude of the initial velocity (Hint:Look at the x direction and solve for Vox)

2. Relevant equations
Other questions I must answer. If you have time, help with these would be great, however I think if I can just find the initial velocity I should be fine.
-Calculate the maximum height (***Actually, help with this one would really be nice!)
-Calculate the time it took from launch to land (Easy enough)

3. The attempt at a solution
Well, I looked a a few different things around the internet and this forum, and I found one thing, maybe....
So, I found the equation s=Vo^2/g*sin(2*theta). Using this equation, solving for Vo, I got 27.586 m/s. This seems like a reasonable enough answer, however I have absolutely no idea if it is correct.
I may have approached it completely wrong, I really don't know....

Thanks for the help! This is due tomorrow, and I really don't understand it.

2. Sep 19, 2010

### rock.freak667

Yes that formula is correct. The '85 m' is called the range, essentially how far it travels in the 'x' direction.

Range = v02sin2θ/g

To find the maximum height, answer this, at the highest point, what would be the vertical velocity? (rememeber, after this point, it starts to move downwards)

3. Sep 19, 2010

### ://Justice

Really, so 27.586 m/s would be correct, then, assuming I solved correctly for Vo? Great.

Yes, so the object is at it's maximum height when V=0. V is decelerating at 9.8 m/s due to gravity. Now, I know that the angle, 33 degrees, is somehow involved here, no? But then what would my equation be =/ Umm... sorry, I am having trouble wrapping my head around an equation to use.

4. Sep 19, 2010

### rock.freak667

Right so vertically final velocity is zero. We have the initial velocity as 27.586 m/s at an angle of 33°, so what is the vertical component of this velocity?

We have
1) Final velocity
2) Initial velocity
3)Acceleration
and we want to find 4) distance.

Do you know a kinematic equation (SUVAT equations) that has all these quantities?

5. Sep 19, 2010

### ://Justice

V2=Vo2+2a(x-xo) ??

If V is 0, Vois 27.586, a is g, X is our unknown, and Xo is 0.

So, 0=27.5862+(2)(9.8)(X)
So, 27.5862 / (2)(9.8)= X
Correct?
Then, 38.826 would be the answer. I think I got it. Thanks :D

6. Sep 19, 2010

### rock.freak667

Yes, that would be correct. But your equation should be (for projectile motion)

v2=v02-2g(x-x0)

That will take care of the negative sign when you rearrange and solve for 'x'.

7. Sep 19, 2010

### ://Justice

Ah, oops. Technicalities always seem to mess me up, haha. Thanks. This forum seems really cool. My physics teacher doesn't seem to know how to teach, so I'll probably be here a lot in the future, haha. Thanks again, you really saved me.

8. Sep 19, 2010

### ://Justice

Hmm... do you think I could have help with that last question that was supposed to be easy enough, Calculate the time it took from launch to land? Sorry, but I am a bit stuck.
So, I would use the formula X=Xo+Vot+.5at^2, correct? But then, I get a bit confused...
-.5t^2=Vot-X
divide by t
-.5t=Vo-X
t=-2(Vo-X)
but that is incorrect, I think... SO I dunno... Help, please... Thank you

Oh, wait... Am I maybe just way over-analyzing this? Would it be as simple as Distance/Velocity? Therefore, V=30.197m/s (I made a mistake before. That is the correct V), and D=85m, so 85/30.197=2.815s. But that seems a bit short. The empirical data is 4s (we must find the percent error), that is quite a large percent error...

Last edited: Sep 19, 2010
9. Sep 20, 2010

### rock.freak667

Right. We need to consider vertical motion to get this time.

If it starts at x0=0, then when it lands, the vertical displacement will be x=0. So put x=0 and solve for t.

10. Sep 20, 2010

### ://Justice

Yeah, I got it after a bit. Thanks for all the help. Sure I'll be back soon with more obvious physics questions, haha.