Question

Can someone please help me with this...
A stone is dropped from the edge of a roof, and hits the ground with a velocity of −185 feet per second. Assume the acceleration due to gravity is -32 feet per second squared. How high (in feet) is the roof?

Answer 1

since the stone is dropped it has no initial velocity

Answer 2

the acc dtg, is integrated upwards to velocity:

v(t) = -32t + C; where of course C = 0 since there is no initial velocity

v(t) = -32t; when v(t) = -185, t=?

Answer 3

-185 = -32t

-185/-23 = t

5.78125 = t, this gives us how long it took for the stone to hit the ground

Answer 4

the integration of velocity gives us position:

h(t) = -16t^2 + Vi t + Hi

Vi = initial velocity, =0

and we want to solve for Hi, knowing that at t=5.78125, h(5.78125) = 0


h(t) = -16t^2 + Hi

h(5.78125) = -16(5.78125)^2 + Hi

0 = -16(33.4228515625) + Hi

16(33.4228515625) = Hi

534.765625 = Hi, or initial Height in feet

Answer 5

Thanks!

Answer 6

yep, hope it made sense :)