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A man stands on the roof of a building of height 13.4m and throws a rock with a velocity of magnitude 34.4m/s at an angle of 32.7❝ above the horizontal. You can ignore air resistance.Calculate the maximum height above the roof reached by the rock.

Calculate the magnitude of the velocity of the rock just before it strikes the ground.  Edit Comment  Solutions
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Part A)

To get to the height, apply vf2 = vo2 + 2ad

0 = (34.4)2(sin232.7) + 2(9.8)(d)

d = 17.6 m

Part B)

The x velocity will be 34.4(cos 32.7) = 28.95 m/s

The y component will be from vf2 = vo2 + 2ad

vf2 = 0 + 2(9.8)(31)

vf = 24.65 m/s

The net is from the Pythagorean Theorem

net2 = (24.65)2 + (28.95)2

net = 38.0 m/s

Part C)

The time to climb to 17.6 m is from

vf = vo + at

0 = 34.4(sin 32.7) + (9.8)(t)

t = 1.90 sec

The time to fall the 31 m is from

d = vot + .5at2

31 = 0 + .5(9.8)(t2)

t = 2.52 sec

The total time = 2.52 + 1.90 = 4.41 sec

Finally apply d = vt

d = (34.4)(cos 32.7)(4.41)

d = 127.7 m  Edit Comment  Close Choose An Image
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