Recall that the equation for the horizontal distance h in feet of a projectile with initial velocity v0 and initial angle is given by

Question

Recall that the equation for the horizontal distance h in feet of a projectile with initial velocity v0 and initial angle theta is given by.

a) Assume the initial velocity is . What initial angle will you need to ensure that the horizontal distance will be exactly 100 feet?

b) Assume the initial velocity is . What is the maximum horizontal distance possible and at what angle does this occur?

Answer 1

a. h = [(60)²/16]sinθcosθ = 100

sinθcosθ = 4/9

2sinθcosθ = 8/9

sin(2θ) = 8/9

2θ = Sin^-1(8/9), so θ = 31.3669777°

b. h = [(60)²/16]sinθcosθ = [(60²/32]2sinθcosθ = 112.5sin(2θ)

dh/dθ = 225cos(2θ) = 0

2θ = 90°, so θ = 45°

Maximum distance = 112.5sin90° = 112.5 ft