R(θ) = v0^2 sin 2θ/ 32 .
If v0 = 2900 ft/s, what angle (in degrees) should be chosen for the projectile to hit a target on the ground at 7000 ft from projection ?
The equation tells us the range attained by a projectile when launched from a point with an initial velocity of v0 ft/s and is fired at an angle of θ to the horizontal.
R = 7000 ft
v0 = 2900 ft/s
Using R = v0^2 sin 2θ/ 32
7000 = (2900)^2 sin 2θ/ 32
sin 2θ = 32*7000/(2900)^2 = 0.0266
2θ = 1.52625 degrees (1º 31' 34.5")
θ = 0.76313 degrees (0º 45' 47.25")
This is a very low angle of inclination, and the projectile would reach its target in just under 2 1/2 second.