A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angl

Question

A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58 degrees above the horizontal. The end of the ramp is 1.2 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?
(b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

Please explain step by step... I've tried this problem so many times! Thanks.

Explanation / Answer

Note that the maximum height of a projectile is

H = vo^2 sin^2 (A) / 2g

Thus,

H = 1.598 above the ramp. As the ramp is 1.2 above the ground,

y = 2.80 m   [ANSWER, PART A]

**************

When you reach the highest point, you travelled half the range of your projectile. As

Range = vo^2 sin (2A) / g

Then

Range = 3.995 m

Thus, half the range is

Range / 2 = 2.00 m   [ANSWER, PART B, distance from the ramp]

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