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I wouldn't normally do this, but I can't get an answer any other way.
A shot fired from a gun that is 4 feet high with a muzzle velocity of 1200 feet per second is to hit a target on the ground 3000 feet away. Determine the minimum angle of elevation of the gun.
Here's the equation:
r(t)= Vcos(Θ)ti + (h+Vsin(Θ)t-16t^2)j
I've tried and I always get the wrong answer. I arrive at the simultaneous equations:
3000=1200cos(Θ)t
0=4+1200sin(Θ)t -16t^2
which I then change to
t=3/12cos(Θ)
to
4 +1200sin(Θ)(3/12cos(Θ))-16(3/12cos(Θ))^2 = 4+ 300tan(Θ) - sec(Θ)^2 = 0
sec(Θ)^2 = 1 + tan(Θ)^2 so
3 + 300tan(Θ) - tan(Θ)^2=0
solve for tan(Θ) using polynomial root finder on TI-89
tan (Θ) = 300.01, -.01
arctan (ans) = 89 degrees or -.573 degrees. When I plug both back into the original equation, they fail.
And to the guy saying I need to convert to SI, gravity is -32 ft/s/s.
Also, everybody here is oversimplifying the problem.
Horizontal velocity is a function of the angle of elevation. So when you raise the angle of elevation, the horizontal velocity decreases.
