Posted by **Justin** on Tuesday, October 5, 2010 at 3:28pm.

A skateboarder shoots off a ramp with velocity of 6.5 m/s directed at a 59 degree angle. The end of the ramp is 1.3m above the ground.

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 -
**bobpursley**, Tuesday, October 5, 2010 at 3:43pm
break the initial vertical velocity into vertical, and horizontal components

vertical Viv=6.5sin59

horizontalVih=6.5cos59

at the highest point, the vertical velocity will be zero.

vf=viv-g*t

t=viv/g

now use that time in

h(t)=hi+viv*t-4.9t^2

how far?

distance=vih*t

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