A skateboarder shoots off a ramp with a velocity of 5.0 m/s, directed at an angle of 57° 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?

By using the suvat equations: (a) v 2 = u 2 + 2 a s for the vertical components, so 0 2 = ( 5 s in 57 ) 2 + 2 ∗ − 9.81 ∗ s therefore s=0.896m However, since the ramp is 1.2m above the ground, the skateboarder is 0.896+1.2=2.096m above the ground.
(b) Work out the time when the skateboarder reaches the highest point v=u+at for the vertical components, so 0=5sin57-9.81 t, therefore t=0.427s Now using the horizontal components: s=ut, so s=5cos57 0.427=1.164m

Answered by Abhi5 | 2024-06-10

The skateboarder reaches a maximum height of approximately 2.04 m above the ground and is about 1.05 m horizontally from the end of the ramp when he reaches that height.
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Answered by Abhi5 | 2024-11-04