A skateboarder shoots off a ramp with a velocity of 6.5 m/s, directed at an angl

Question

A skateboarder shoots off a ramp with a velocity of 6.5 m/s, directed at an angle of 59° above the horizontal. The end of the ramp is 1.4 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp.

(a) How high above the ground is the highest point that the skateboarder reaches?
m

(b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
m

Explanation / Answer

Part A)

We need the additional height

Apply vf2 = vo2 + 2ad

vo = 6.5sin59 in the y direction

That is 5.57 m/s

0 = 5.572 + 2(9.8)(d)

d = 1.58 m

1.58 + 1.4 = 2.98 m

Part B)

Time to reach that height

vf = vo + at

0 = 5.57 + (9.8)t

t = .568 sec

The x distance is from d = vt

d = (6.5)(cos 59)(.568)

d = 1.90 m

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