Question

The drawing shows a skateboarder moving at 4.8 m/s along a horizontal section of a track that is slanted upward by 48° above the horizontal at its end, which is 0.50 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

Let M = mass of the skier, 
v2 = his speed at the end of the track. 
By conservation of energy, 
1/2 Mv^2 = 1/2 Mv2^2 + Mgh 
Dividing by M, 
1/2 v^2 = 1/2 v2^2 + gh
 Multiplying by 2, 
v^2 = v2^2 + 2gh 
Or v2^2 = v^2 - 2gh 
Or v2^2 = 4.8^2 - 2 * 9.8 * 0.46 
Or v2^2 = 23.04 - 9.016 
Or v2^2 = 14.024 m^2/s^2-----------------------------(1) 
In projectile motion, launch speed = v2 
and launch angle theta = 48 deg 
Maximum height 
H = v2^2 sin^2(theta)/(2g) 
Substituting theta = 48 deg and value of v2^2 from (1),
 H = 14.024 * sin^2(48 deg)/(2 * 9.8) 
Or H = 14.024 * 0.7431^2/19.6 
Or H = 14.024 * 0.5523/19.6 
Or H = 0.395 m = 0.4 m after rounding off 
Ans: 0.4 m

The answer in this question is 0.4 m