trig question please help

Question

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 7000 ft

Answer 1

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.