A person standing on top of a 40-m high cliff throws a rock with a mass of 0.25 kg off the cliff.The rock leaves the person's hand with a speed of 11 m/s and makes an angle of 75 degrees above the horizontal.What is the speed of the rock when it hits the ground at the bottom of the cliff?

Question

Quizplus · Accepted Answer

We can use the conservation of energy to find the velocity of the rock at the bottom of the cliff. 
Initial Potential Energy + Initial Kinetic Energy = Final Potential Energy + Final Kinetic Energy

(0.25kg)(9.81m/s^2)(40m) + 0.5(0.25kg)(11m/s)^2 = (0.25kg)(9.81m/s^2)(0m) + 0.5(0.25kg)(v_final)^2

Simplifying, we get:

98.1J + 15.125J = 0J + 0.125v_final^2

Solving for v_final, we get:

v_final = sqrt((98.1J + 15.125J) / 0.125(0.25kg)) = 30 m/s

Therefore, the answer is C.

A Person Standing on Top of a 40-M High Cliff