A stone is thrown straight up from the edge of a roof, 900 feet above the ground, at a speed of 10 feet per second.

Question

A. Remembering that the acceleration due to gravity 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

h(t) = -16t^2 +10t + 900

height after 2 seconds =

h(2) = -16(2)^2 +10(2) + 900

=-64 +20 + 900

= 856 feet after 2 seconds

h(t) = -16t^2 +10t + 900 = 0 when it hits the ground, divide by -1

16t^2 - 10t - 900 = 0 divide by 2

8t^2 -5t - 450 = 0, use the quadratic formula to solve for t

t = 5/16 + (1/16)sqr(25^2 +32(450)) = slightly less than 8 seconds to hit the ground

velocity is the derivative of the height function

v(t) = h'(t) = -32t + 10

plug in t=8

v(8) = -32(8)+10 = -256+10 = about -246 feet per second = velocity when it hits the ground