A man standing on top of a building throws a ball up into the air, releasing it from a height of exactly 48 feet with an initial velocity of 32 feet per second.

Question

A man standing on top of a building throws a ball up into the air, releasing it from a height of exactly 48 feet with an initial velocity of 32 feet per second. How long will it take the ball to hit the ground?

Answer 1

The general equation is:

h(t) = -16t² + v₀t + h₀

where:

1. h(t) = height at time t
2. t = time in seconds
3. v₀ = initial velocity = 32 f/s
4. h₀ = initial height = 48 f

h(t) = -16t² + 32t + 48

The height when it hits the ground is 0:

0 = -16t² + 32t + 48

Divide both side s by -16:

0 = t² - 2t - 3

Solve for t by factoring (it does factor) or using the quadratic formula. You'll get two answers, one positive and one negative. Since I haven't quite perfected my time machine, the positive answer is the one you want. Units are seconds.