A stone is thrown at an angle of 30° above the horizontal from the top edge of a cliff with an initial speed of 14 m/s.A stop watch measures the stone's trajectory time from top of cliff to bottom to be 7.1 s.How far out from the cliff's edge does the stone travel horizontally? Note: g = 9.8 m/s² and air resistance is negligible. A) 58 m B) 68 m C) 86 m D) 197 m

Question

Quizplus · Accepted Answer

Using the kinematic equations: 
y = voy*t + 1/2*a*t^2, where y is displacement in the y direction, voy is initial velocity in the y direction, a is acceleration in the y direction (due to gravity), and t is time.

Using angle 30°, the initial velocity can be calculated as V0x = V0*cos(30°) = 12.124 m/s and V0y = V0*sin(30°) = 7 m/s.
y = V0y*t + 1/2*a*t^2
0 = 7*t - 1/2*9.8*t^2
t = 1.43 s

Using the same initial velocity, the horizontal displacement can be calculated as x = V0x*t.
x = 12.124*7.1
x = 86 m

A Stone Is Thrown at an Angle of 30° Above