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