In: Physics
A man stands on the roof of a building of height 15.7 m and throws a rock with a velocity of magnitude 30.1 m/s at an angle of 25.0 ? above the horizontal. You can ignore air resistance.
A) Calculate the maximum height above the roof reached by the rock
B)Calculate the magnitude of the velocity of the rock just before it strikes the ground
C) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.
Height h = 15.7 m
Initial velocity v = 30.1m/s
Angle = 25.0 o
(a). maximum height reached by the rock above he roof is H = ( v sin ) 2 / 2g
H = (30.1 sin 25.0 ) 2 /(2 *9.8)
= 8.25 m
(b).Horizontal velocity when strikes the ground = v cos 25.0
= 30.1 cos 25.0
v ' = 27.28 m/s
Vertical velocity when it strikes the ground V = ?
Initial velocity in vertical direction u = v sin 25.0
= 30.1* sin 25.0
= 12.72 m/s
Accleration a = g = 9.8m/s 2
From the relation V 2 - u 2 = 2aS
V 2 = u 2+ 2aS where S = h
= 12.72 2 +(2 *9.8* 15.7)
= 469.5
V = 21.7m/s
Magnitude of the velocity of the rock just before hit the ground = [v ' 2 + V 2]
= [27.28 2 +21.72]
= 34.86 m/s
(C).Horizontal distance R = v ' x t
Where t = time taken to hit the ground
From the relation V = u + at
at = V - u
= 21.7-12.72
= 8.98
t = 8.98 / a
= 8.98 / g = 8.98 / 9.8
= 0.9163s
So, R = 27.28*0.9163
= 24.99m