The Drawing Shows A Skateboarder Moving At 5.4 M/s

* 43. � ssm The drawing shows a skateboarder moving at 5.4 m/s along a horizontal section of a track that is slanted upward by 48� above the horizontal at its end, which is 0.40 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.

First break this up into two parts.� Part A is along the track.� Part B is leaving the track and being a projectile.� Along the track, we are ignoring friction and air resistance so we have no non-conservative forces, so along the track mechanical energy is conserved.� So we can find the speed of the skater as she leaves the track.� Let�s call this v₁

Since we will make h₀ = 0 (on the ground)� So solve for v₁

Now to find the height above the launch point, we can use conservation of energy again.� For this part, make the top of the track a new zero of gravitational potential energy.� Again with no air resistance we have a conservation of mechanical energy.

Remember h₁ is now zero and h₂ is H.� So solve for H

Now the speed at the top of the arc is the horizontal speed only as the skater goes up until the vertical component of the speed has gone to zero.� So we know that v₂ is the horizontal part of v₁.