A particle is launched from the top of a cliff of height 87.5m at time t=0 with initial velocity 14m/s at 30 deg above the horizontal, Calculate: a) maximum height reached above bottom of cliff; b)horizontal distance travelled before hitting the ground.

For parabolic motion: vertical and horizontal components can be treated independently. x represents horizontal motion, which obeys constant velocity equations, y denotes vertical motion, which obeys the constant acceleration suvat equations, with a equal to g. For this question, the upward direction is taken as positive.

a) Max height: Use s to denote the height reached by the particle above the launch point, we can use the equation v²=u² + 2as. At its maximum height, the vertical velocity of the particle, v_y=0. u=u_y=u sin 30 = 14sin30 = 7m/s and a= -9.8m/s². Rearranging the equation gives s=(v²-u²)/2a = (0-(7)²) / 2(-9.8)= -49 / -19.6 = 5/2 = 2.5m. However, we want height relative to the bottom of the cliff, so we must add 87.5m. Therefore, the maximum height reached is 2.5 +87.5 =90m

b) Horizontal range: s_y = vertical change in displacement relative to launch point = drop in height of cliff height = -87.5m, u_y = 14sin30 = 7m/s, a= -9.8m/s². Using s=ut +(at²)/2 and substituting in the given values,  gives -87.5=7t-4.9t², rearranged to give 4.9t²-7t-87.5=0. This can be solved to give t=5s as t is greater than 0. To find the range, the horizontal distance must be calculated. For constant velocity motion: velocity=(change in displacement)/time, or v_x = s_x/t. Rearranging gives s_x=v_x t. v_x = u cos30 = 14 cos30 m/s and t=5s, so s_x=5 x 14 cos30 = 70cos30 = (70/2) sqrt(3) = 35 sqrt(3) = 60.6m