Posted by **Elle** on Friday, February 21, 2014 at 6:59pm.

A skateboarder shoots off a ramp with a velocity of 5.7 m/s, directed at an angle of 58° above the horizontal. The end of the ramp is 1.3 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?

- Physics -
**Damon**, Friday, February 21, 2014 at 7:24pm
u = horizontal velocity = 5.7 cos 58

= 3.02 m/s forever

Vi = initial speed up = 5.7 sin 58

= 4.83 m/s

v = Vi - 9.8 t

at top, v = o

0 = 4.83 - 9.8 t

t = .493 s at top

h = 1.3 + Vi t - 4.9 t^2

t at top is .493

so

h at top = 1.3 + 4.83(.493)-4.9(.493)^2

= 2.49 meters above ground

x = u t

x = 3.02 * .493 = 1.49 meters

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