physics word problems

Question

The drawing shows a skateboarder moving at 7.10 m/s along a horizontal section of a track that is slanted upward by 0 = 42.0° above the horizontal at its end, which is 0.740 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.

Answer 1

Denote V_t as the speed at the end of the track.

L is the length of the track and L = 0.74/sin42

Once on the slope, the skater will be under a constant deceleration of g*sin42 down along the slope.

Therefore, V_t² = V₀²-2L*g*sin42=7.1²-2*0.74*g =36

Note that sin42 is cancelled, the angle doesn't matter for its ending speed.

So V_t=sqrt(36)=6m/s. This speed is 42 degree upward, the vertical speed going up would be Vt*sin42 =4m/s.

Using V²=2gh, 4²=2*10*h, h=0.8(m), roughly.