Projectile Motion question?

Question

I'd just like some reasoning and an explanation on how to do this question, you don't necessarily have to do it.

A hammer at the top of a roof slides from rest and falls to the ground at P. The roof is smooth and sloped at an angle of 30 degrees to the horizontal as shown in the figure below:
http://i242.photobucket.com/albums/ff10/breakthiscycle/101-1.jpg
The roof is 10 meters long and its lowest point is 10 meters from the ground. Calculate:
a)to the nearest significant figure, the velocity of the hammer when it reaches the lowest point on the roof.
b)the horizontal component of its velocity when it strikes the ground.
c)the vertical component of its velocity when it strikes the ground
d)the angle that the velocity of the hammer makes with the ground.

Answer 1

a) v^2 = 2gh ; h = 10 + 10 sin 30 = 15 m
v = 17.15m/s
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b) Horizontal component at the ground is the horizontal component when it just leaves the roof.

Use v^2 = 2as where a = g sin 30, s = 10

v = 9.9m/s

The horizontal component is 9.9 sin 30 m/s = 4.95m/s
The vertical component is 9.9 cos 30 m/s = 8.57m/s.

Horizontal component at the ground is also 4.95m/s
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c)

The vertical component 8.57m increases from this value when it reaches the ground.

v^2 = u^2 + 2gh .

v = 16.41 m/s
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d)

The angle is found using tan θ = horizontal component/ vertical component/

tan θ = 4.95/ 16.41

θ = 16.78 from the vertical direction.
Or

tan β = 16.41 / 4.95]

β = 73.22 from the ground.
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Check the result in a)

√[16.41^2 + 4.95^2 ]== 17.14 m/s

Answer 2

With no friction, the acceleration on the hammer is
a = gsin30°
The velocity of the hammer when it reaches the edge of the roof will be
v1 = (2*10gsin30°)^(1/2)
From this point until it reaches the ground, the horizontal velocity (v1x = v1cos30°) will remain unchanged.
The vertical component of velocity when it strikes the ground will be
v2y = [(v1sin30°)^2 + 2*10g]
d)the angle that the velocity of the hammer makes with the ground is tan^-1(v2y/v1x)

Answer 3

a) You have to use the equation of motion v^2=u^2+2as(1) to calculate the velocity at the lowest point on the roof.

b) At the lowest point of the roof, the horizontal component is vcos30, then you just use equation (1) to find the velocity on the ground.

c) Same with (b), but the vertical component is vsin30 at the lowest point of the roof.

d) You have to use conservation of energy to find its velocity on the ground, then from its vertical and horizontal component found in (b) and (c), you can determine the angle it makes with the ground.

Answer 4

If we can ignore the friction of the roof and air resistance, the problem breaks into two parts.
The slide will produce a velocity due to gravity(a), the vertical component will be V*sine 30 and the horizontal (b)will be V*cos 30.
(c)The fall is basic gravity acceleration with an initial velocity of V*sin30
The horizontal V remains the same from roof edge to ground.
(d) can be calculated by Pythagoras's Theorem using final vert. velocity as the adjacent side and the horizontal velocity as the opposite side.