A stone is dropped from the edge of a roof, and hits the ground with a velocity of -125 feet per second. How high (in feet) is the roof?

Question

Note: the acceleration of an object due to earth's gravity is 32 ft/sec²

Answer 1

y = h - gt²/2 = 0; h = gt²/2; v(t) = dy/dt = - gt; t = - v(t)/g; h = g/2·v²(t)/g² = v²(t)/(2g) = 125²ft²/s²/(2·32ft/s²) =

244.14 ft

Answer 2

h(t) = -16t^2 + ho where ho = initial height = height of the building, and initial velocity = 0

take the derivative for velocity and set = velocity when it hits the ground, solve for t

v(t) = h'(t) = -32t = -125

t = 125/32 = about 3.9 seconds to hit the ground, plug that value into the h(t) function

h(t) = -16t^2 + ho = 0 (h = 0 for ground level)

h(125/32) = -16(125/32)^2 = -ho

ho = 15625/64 = 244.140625 feet = about 244 feet = height of the building's roof