Calculating the Impact of a Hammer on a Lawn

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The problem involves calculating the distance a hammer will land from a wall after sliding down a roof inclined at 60 degrees and falling 11 meters to the lawn. The net force acting on the hammer is determined to be 7.36 N, leading to an acceleration of 7.36 m/s². The time taken for the hammer to fall 11 meters is calculated to be approximately 0.604 seconds. Using this time, the horizontal distance from the wall is found to be 5.3 meters. The calculations appear to be correct, despite initial doubts.
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Homework Statement



A hammer slides 21 m down the side of the roof inclined at 60 deg. above the horizontal, and falls down to the lawn 11m below the edge of the roof.

How far away from the wall will this hammer impact with the lawn?

m = 1 kg
mu = 0.23


Homework Equations





The Attempt at a Solution



Fgx = mg sin 60
= (1) (9.8)sin60
= 8.487N

Ff = UFn = 1.127 N

Fnet = Fgx - Ff = 8.487 - 1.127 = 7.36 N

Fnet = ma
a = Fnet/m = 7.36 m/s^2

to find t sub y = 11

y = yo + voyt - 1/2gt^2
0 = 4.9t^2 + 15.23t - 11
t = -15.23 +- (15.23^2 - 4(4.9) (-11)))^sq root

t = -15.23 +- (15.23^2 - 4(4.9) (-11)))^sq root / 2 (4.9)

t = -15.23+- (447.5529)^1/2/ 9.8

then t = 0.06004

sub t = 0.604 into x = Vxt
x = 17.582cos60 (0.604)
x = 5.3 m

5.3 m away from the wall

?? i rly doubt this is rite...??
 
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It appears to be correct.
 
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