Posted by **Stuck** on Thursday, July 29, 2010 at 8:45pm.

"A workman sitting on top of a roof of a house drops his hammer. The roof is smooth and slopes at an angle of 30.0 degrees to the horizon. It is 32m long and its lowest point is 32m from the ground. How far from the house's wall is the hammer when it hits the ground?" (28m)

I thought I would firstly find the acceleration of the hammer down the roof.

ma = Fgx

a = (9.81)(sin30) = 4.905

Using the acceleration, I found the final velocity at the roof.

Vf^2 = Vi^2 + 2ad

Vf^2 = 0 + 2(4.905)(32)

However, is the initial velocity 0?

After I found Vf, I used that as the initial velocity to find the time it would take for the hammer to fall to the ground, and then used that time to find the horizontal distance. I don't get the answer though...am I even doing this correctly?

- Physics -
**bobpursley**, Thursday, July 29, 2010 at 9:13pm
Your Vf (velocity at the edge) is correct. However, now you have to break that into a horizonal velocity, and a vertical velocity.

Verticalatroofedge=sqrt (2*4.9*32)

horizontal component=sin30*sqrt( )

vertical component= cos30*sqrt( )

now you can find the time down...

32=verticalcompnent*t+1/2 g t^2 solve for t.

then, horizonal distance= vhorizontal component*t

- Physics -
**Damon**, Thursday, July 29, 2010 at 9:17pm
I would use energy to get the speed at the roof edge.

(1/2) m v^2 = m g * 32 sin 30

v^2 = 9.81 * 64 (.5) = 313.9

v = 17.7 m/s which agrees with you so far

However now you have two problems, a horizontal problem and a vertical problem.

The horizontal speed is 17.7 cos 30 = 15.3 m/s and remains constant

The initial vertical speed down is 17.7 sin 30 = 8.85

so

32 = 8.85 t + .5*9.81 t^2

solve that for t

then distance horizontal = 8.85 t

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