Question

An rock is thrown upward from a platform that is 188 feet above ground at 80 feet per second. Use the projectile formula to determine when the rock hit the ground.

Answer 1

6.74 s

Explanation:

y = ½ at² + vt + h

Given a = -32, v = 80, h = 188:

y = ½ (-32) t² + (80) t + (188)

y = -16t² + 80t + 188

When y = 0:

0 = -16t² + 80t + 188

0 = 4t² − 20t − 47

Solve with quadratic formula:

t = [ -(-20) ± √((-20)² − 4(4)(-47)) ] / 2(4)

t = [ 20 ± √(400 + 752) ] / 8

t = (20 ± √1152) / 8

t > 0, so:

t = (20 + √1152) / 8

t ≈ 6.74