Answer to Question #166180 in Physics for gibson mkandawire

Question

A stone is thrown horizontally from a cliff 30m high. The initial velocity is 6m/s.

(a) Sketch the motion of the stone.

(b) How long is the stone in air?

(c) What is the velocity of a stone as it hits the ground?

(d) At what angle does stone strike the ground?

(e) At what distance from the cliff base does the stone strike the ground?

Answer 1

(a) Let's sketch the motion of the stone:

(b) We can find the flight time of the stone as follows:

$$y=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2y}{g}}=\sqrt{\dfrac{2\cdot30\ m}{9.8\ \dfrac{m}{s^2}}}=2.47\ s.$$

(c) The horizontal component of the stone's velocity remains unchanged during the flight and equals $6\ \dfrac{m}{s}$. Let's find the vertical component of the stone's velocity:

$$v_y=v_{0y}-gt=0-9.8\ \dfrac{m}{s^2}\cdot2.47\ s=-24.21\ \dfrac{m}{s}.$$

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

Finally we can find the velocity at which the stone hits the ground from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(6\ \dfrac{m}{s})^2+(-24.21\ \dfrac{m}{s})^2}=24.94\ \dfrac{m}{s}.$$

(d) We can find the angle at which the stone strikes the ground from the geometry:

$$\theta=sin^{-1}(\dfrac{v_y}{v})=cos^{-1}(\dfrac{-24.21\ \dfrac{m}{s}}{24.94\ \dfrac{m}{s}})=-76.1^{\circ}.$$

The sign minus means that the final velocity of the stone as it hits the ground directed at an angle $76.1^{\circ}$ below the ground.

(e) We can find the distance from the cliff base to the place where the stone strikes the ground from the kinematic equation:

$$x=v_0t=6\ \dfrac{m}{s}\cdot2.47\ s=14.82\ m.$$