Answer to Question #192018 in Field Theory for Pranali Patil

Question

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18m/s.  The cliff is 50.m above a flat horizontal beach.  

How long after being released does the stone strike the beach below the cliff?  

With what velocity does it land?

Answer 1

(a) We can find the time that the stone takes to reach the beach below the cliff from the kinematic equation:

$$y=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2y}{g}},$$ $$t=\sqrt{\dfrac{2\cdot50\ m}{9.8\ \dfrac{m}{s^2}}}=3.19\ s.$$

(b) Let's first find the vertical component of the stone's velocity:

$$v_y=-gt=-9.8\cdot3.19\ s=-31.3\ \dfrac{m}{s}.$$

The sign minus means that the vertical velocity of the stone directed downward.

Finally, we can find with what velocity the stone lands from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2},$$ $$v=\sqrt{(18.0\ \dfrac{m}{s})^2+(-31.3\ \dfrac{m}{s})^2}=36.1\ \dfrac{m}{s}.$$