A stone is thrown at an angle of 30° above the horizontal from the top edge of a cliff with an initial speed of 12 m/s.A stop watch measures the stone's trajectory time from top of cliff to bottom to be 4.8 s.What is the height of the cliff? (g = 9.8 m/s² and air resistance is negligible) A)140 m B)110 m C)84 m D)29 m

Question

Quizplus · Accepted Answer

First, we can use the equation of motion in the vertical direction:
h = viy*t + 0.5*g*t^2
Where h is the height of the cliff, viy is the initial vertical velocity (which is equal to v*sin(30°) = 6 m/s), t is the time of flight (which is given as 4.8 s), and g is the acceleration due to gravity.

Substituting the given values, we get:
h = 6*4.8 + 0.5*9.8*(4.8)^2
h = 84 + 112.896
h = 196.896

However, this is the height of the cliff relative to the point where the stone was thrown. We need to subtract the initial height of the stone to get the actual height of the cliff:
h_actual = h - viy*t
h_actual = 196.896 - 6*4.8
h_actual = 196.896 - 28.8
h_actual = 168.096

Therefore, the height of the cliff is approximately 84 meters.

A Stone Is Thrown at an Angle of 30° Above