a skateboarder moving at v = 5 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is h = 0.52 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track.
Start at the bottom of the track. Her verical component of velocity is 5sinTheta. So the KE which drives the vertical rise is 1/2 m 25sin^2 theta.
Set that equal to the change in potential energy, mgh. Solve for h , the height above GROUND. To get the heighht above the end of track,subtract 0.52m.
I'm sorry I'm not very good at this but can you please clarify:
1. 5sinTheta - are you saying the sign of 5 degrees and what do I do with it?
2. 1/2m25sin^2theta - are you saying 1/2mv(squared) then find the sine of v (squared)?
Thanks!
"5 sin theta" means 5 x sin 48 = 3.716
"theta" is the slope angle.
"1/2m25sin^2theta " means
(1/2)x m x 25 x (sin 38)^2
(The 5-squared became the 25)
You don't need to know tha mass m to solve the problem, because it will cancel out when potential energy m g h is set equal to (1/2)x m x 25 x (sin 38)^2.
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