A projectile fired from ground level is to reach a maximum height of 1600 feet. What is its initial velocity?

Answer 1

To find the initial velocity of the projectile, we can use the kinematic equation for the maximum height of a projectile:

H = (v^2 * sin^2(theta)) / (2 * g)

where:
H = maximum height (1600 feet)
v = initial velocity
theta = launch angle (assuming 45 degrees for simplicity)
g = acceleration due to gravity (32.2 feet/second^2)

Rearranging the equation, we have:

v^2 = (2 * g * H) / sin^2(theta)

Substituting the given values:

v^2 = (2 * 32.2 * 1600) / sin^2(45)

v^2 = 103040 / (0.5)

v^2 = 206080

v = sqrt(206080)

v ≈ 453.7 feet/second

Therefore, the initial velocity of the projectile is approximately 453.7 feet/second.