Question

Assume air resistance is negligible and gravitational acceleration is 32.2 ft/s^2, a projectile is launched at 52 degrees above the horizontal with an initial velocity of 30 ft/s. The launch and landing sites are at the same elevation. What is the projectile's range?

A. 22.0 ft
B. 72.7 ft
C. 27.1 ft
D. 42.0 ft

Answer 1

The range of the projectile can be calculated using the formula:

R = (v^2/g) * sin(2θ)

where v is the initial velocity, g is the gravitational acceleration, θ is the launch angle.

Substituting the given values:

R = (30^2/32.2) * sin(2*52)

R = 72.7 ft

Therefore, the answer is B. 72.7 ft.