*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.

The height H can now be written

 

 

 

 

Alternatively we could switch to kinematics

 

 

 

 

 

 

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This page last updated on January 11, 2020