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.
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
A vector quantity is a quantity which has both magnitude and direction. Here in the given options, speed is a scalar quantity but not the vector quantity.
