Southern Union
posted by Mary Anne .
A large building has an inclined roof. The length of the roof is 64.0 m and the angle of the roof is 20.0° below horizontal. A worker on the roof lets go of a hammer from the peak of the roof. Starting from rest, it slides down the entire length of the roof with a constant acceleration of 3.35 m/s2. After leaving the edge of the roof, it falls a vertical distance of 33.5 m before hitting the ground. (a) How much time does it take the hammer to fall from the edge of the roof to the ground? (b) How far horizontally does the hammer travel from the edge of the roof until it hits the ground?

How fast is it going when it leaves the roof?
v = a T
d = .5 a t^2
64 = .5 * 3.35 * t^2
so t = 6.18 seconds on roof
v = 3.35 * 6.18 = 20.7 m/s leaving roof
Vertical velocity leaving roof = 20.7 sin 20 deg = 7.08 m/s
horizontal velocity roof to ground = 20.7 cos 20 = 19.5 m/s
do vertical problem first
Vi = 20.7
a = 9.8 so a/2 = 4.9
Hi = 33.5
0 = 33.5  20.7 t  4.9 t^2
so
t^2 + 4.22 t  6.84 = 0
t = [ 4.22 +/ 6.72 ]/ 2
= 1.25 seconds
distance horizontal = 19.5 t = 19.5*1.25 = 24.4 metars