Question

A large building has an inclined roof. The length of the roof is 56.5 m and the angle of the roof is 17.0° 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 2.87 m/s². After leaving the edge of the roof, it falls a vertical distance of 22.5 m before hitting the ground.

(a)How much time, in seconds, does it take the hammer to fall from the edge of the roof to the ground?

s

(b)How far horizontally, in meters, does the hammer travel from the edge of the roof until it hits the ground?

m

Answer 1

from equation of v^2 - vo^2 = 2*a*s

v^2 - 0^2 = 2*2.87*56.5

v = 18 m/s

speed of the hammer at the dge of roof v = 18 m/s

along vertical

initial velocity voy = -18*sin17 = -5.26 m/s

acceleration ay = -g = -9.8 m/s^2

displacement y = -22.5 m

y = voy*t + (1/2)*ay*t^2

-22.5 = 0 -(1/2)*9.8*t^2

t = 21.4 s

=================

part(b)

along horizontal

initial velocity vox = v*costheta

acceleration ax = 0

x = vox*t + (1/2)*ax*t^2

x = 18*cos17*2.14

x = 36.8 m <<<<<-------------ANSWER