# A shot fired from gun, muzzle velocity of 1200 ft/s

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1. Feb 14, 2016

### lightningkid99

• Mentor note: moved to homework section
A shot fired from a gun with a muzzle velocity of 1200 ft per second is to hit a target 3000 feet away. Determine the minimum angle of elevation of the gun?

Ok so I know that r(t)= (vcos(theta) t)i +(vsin(theta) t +16t^2)j

I tried to set vcos(theta)t =3000, solve for t, and plug into the j component. However, I am getting confused with the math because i get t= 5/2cos(theta) and when I plug it in it gets pretty messed up. Can someone help?

2. Feb 14, 2016

### Staff: Mentor

It should not get too messy. Please show your calculations, otherwise it is impossible to tell what went wrong.
You either have a strange sign or angle convention with the j component.

3. Feb 14, 2016

### lightningkid99

1200cos(theta) t=3000
t=3000/1200cos(theta) or 5/2cos(theta)

so I plug into the j component
1200sin(theta)(3000/1200cos(theta))- 16(5/2cos(theta))^2

That's where I'm stuck. I'm really not sure where to go from here

4. Feb 14, 2016

### Staff: Mentor

There are brackets missing for denominators, and the last expression should be equal to something. That allows to simplify it.