Answer to Question #164961 in Physics for Jason

Question

Suppose a person is standing on a 20 meter cliff in West Texas. The person picks up a 1.5 kg rock; and with the help of a slingshot, slings the rock with a velocity of 12 m/s. 

1. What is the kinetic energy and potential energy of the rock right after the rock is released from the slingshot?

1. With what velocity will the rock hit the ground at the bottom of the cliff?

Answer 1

1) Right after the rock is released from the slingshot, all the potential energy of the rock stored in the slingshot converted into the kinetic energy:

$$PE=KE=\dfrac{1}{2}mv^2=\dfrac{1}{2}\cdot1.5\ kg\cdot(12\ \dfrac{m}{s})^2=108\ J.$$

2) Let's first find the time that the rock takes to reach the bottom of the cliff:

$$h=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\cdot20\ m}{9.8\ \dfrac{m}{s^2}}}=2.02\ s.$$

The horizontal velocity of the rock equals 12ms^-1 and remains unchanged during the flight. Let's find the vertical velocity of the rock right before the rock hits the ground:

$$v_y=v_{0y}-gt=0-9.8\ \dfrac{m}{s^2}\cdot2.02\ s=-19.8\ \dfrac{m}{s}.$$

Finally, we can find velocity of the rock right before it hits the ground from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(12.0\ \dfrac{m}{s})^2+(-19.8\ \dfrac{m}{s})^2}=23.15\ \dfrac{m}{s}.$$