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I need help with a physics problem on projectile motion

S

Swanny

Diamond Member
Hello all,

I'm doing a physics assignment right now. I'm done with all the other problems, but this one has me stuck. Here it goes: On ground level, what is the maximum range of a cannon with a muzzle velocity of 336 m/s?

I know this must have something to do with finding the best angle of elevation for the cannon. I can just see the graph on a graphing calculator, but I don't have one at home. Can someone tell me how to solve this longhand?


Thanks,
Swan
 
High school was a LONG time ago for me, but I might be able to help a little bit.

Combining my high school physics and my two seasons on the shotput team, the best angle for horizontal range is usually under 45 degrees and more than 26 degrees. If I remember correctly, the coach always wanted 35 to 40 degrees, so assuming the laws of physics are roughly the same for a cannon, I'd try in that range first. 🙂

I'd be interested to hear the answer too.
 
Since I'm assuming this is a frictionless environment, to get maximum range you want to fire at 45 degrees, so you'll have the best combination of hangtime and forward velocity.
To solve, you need to find out how long the projectile will stay in the air if the initial y velocity is 336 * sin(45). After you get that time, multiply it by the x velocity (336 * cos(45)) to get the max range.
 
Or you can use range = ((vi^2) * sin(2*theta))/g

Thus you'll see that sin (2*45) = 1 will provide the maximum value for range.

I've tried both my method and the regular solve for time and plug back into s(xcomponent) = V(xcomponent) *t and both were equal.
 
Alright, it's been like 5 years since high school physics, but... this is what I've come up with. No idea if it's right.

I think you have to break your initial velocity vector into its component parts... so
336^2 = (h-velocity)^2 + (v-velocity)^2 OR
112896 = (h-velocity)^2 + (v-velocity)^2

When the cannon ball hits the ground again, its horizontal velocity will be the negative of its initial horizontal velocity:
(v-velocity) - 9.8t = -(v-velocity) OR
2(v-velocity) = 9.8t

The distance travelled, which you want to maximize, is a function of your initial horizontal velocity:
d = (h-velocity)*t


Now if you solve the vertical velocity equation for t, and substitute that for t in the horizontal velocity equation, you'd get:
d = ((h-velocity)*(v-velocity)) / 4.9

To maximize d, h-velocity and v-velocity would have to be the same, right? So the ideal angle would be 45 degrees.

Don't trust me on that though, I could be totally wrong. I used to be good at math...

Edit: ok fine, I tried doing it without sines because I didn't have a calculator. 🙂 I got the right answer though!
 
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