Hello, could someone help me set up the equation for this word problem?
A sharpshooter intends to hit a target at a distance of 3000ft with a gun that has a muzzle velocity of 1200 ft per second. Neglecting air resistance, determine the gun's minimum angle of elevation theta if the range r is given by r=1/32v^2/0sin(2theta). Round your angle to nearest tenth of a degree.
well, geez, just solve the equation.
r = v^2sin2θ/g
1/32 1200^2sin(2θ) = 3000
sin2θ = (32*3000)/1200^2 = 1/15
Of course, there are two solutions, since sinθ is positive in [0,pi]. But you want the minimum value.