Question

A stone is thrown straight up from the edge of a roof, 675 feet above the ground, at a speed of 14 feet per second. A. Remembering that the acceleration due to gravitv is −32 feet per second squared, how high is the stone 2 seconds later? B. At what time does the stone hit the ground? C. What is the velocity of the stone when it hits the ground?

Answer 1

Step-by-step explanation:

A.) d = di + Vit + 1/2at²

d = 675 + (14 ft/s)(2 s) + 1/2(-32 ft/s²)(2)² = 639 ft

Stone goes up to a max. height, then starts to fall down. At t = 2 s, the stone is 639 ft high.

B.) When the stone hits the ground, d = 0

0 = 1/2(-32 m/s²)t² + (14 ft/s)t + 675

Solve for t using quadratic equation: a=-16, b=14, c=675

t = -6.07, 6.95 disregard the negative root

t = 6.95 s when stone hits the ground

C.) vf = vi + at

vf = 14 m/s + (-32 ft/s²)(6.95 s) = -208.4 ft/s (down)