A stone is thrown at an angle of 30 $\degree$ 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 5.6 s.What is the height of the cliff? (g = 9.8 m/s² and air resistance is negligible) A)58 m B)154 m C)120 m D)197 m

Question

Quizplus · Accepted Answer

We can use the kinematic equations of motion to solve this problem.

First, we need to find the vertical component of the initial velocity:

v0y = v0sinθ = 12sin30° = 6 m/s

Using the kinematic equation:

Δy = v0yt + 0.5at^2

where Δy is the height of the cliff, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight (5.6 s), we can solve for Δy:

Δy = v0yt + 0.5at^2 = 6(5.6) + 0.5(-9.8)(5.6)^2 = 120.96 m

So the height of the cliff is approximately 120 m (to two significant figures). Therefore, the correct answer is C.

A Stone Is Thrown at an Angle of 30 °\degree° Above the Horizontal from the Top Edge of a Cliff

A Stone Is Thrown at an Angle of 30 $\degree$ Above the Horizontal from the Top Edge of a Cliff