A stone is thrown straight up from the edge of a roof, 775 feet above the ground, at a speed of 16 feet per second.

Question

A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 3 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?

please do it step by step

Answer 1

A) Use Δy = v_o t + (1/2) a t² = 16*3 + (1/2)(-32)*9 at 3 seconds, Δy = -96 ft and y = 775 + Δy = 679 ft.

B) Same equation, -775 = 16 t - 16 t², 16 t² - 16 t - 775 = 0. Apply quadratic formula and choose positive root: t = {16 ± √[256 - 4(16)(-775)] } /32 --> [16 + 223]/32, t = 7.48s.

C) Use v² = v_o² + 2a Δy = 16² + 2(-32)(-775) --> v = 223 ft/s.