Answer to Question #254672 in Mechanics | Relativity for gold

Question

1. A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above a flat, horizontal beach. Refer to the figure on the module posted. 

A. What are the components of the initial velocity?

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

C. With what speed and angle of impact does the stone land?

Answer 1

A. Initial velocity is in horizontal direction

x-component = 18 m/s

y-component = 0

B.

$50 = 0 + \frac{1}{2} \times g \times t^2 \\ t = \sqrt{\frac{2 \times 50}{9.8}} = 3.1943 \;s$

C.

$V_y =(2gh)^{0.5} = 31.30495 \; m/s \\ V =(V_x^2 +V_y^2)^{0.5} = 36.1109 \; m/s \\ Angle = arctan( \frac{V_y}{V_x)} =60.11 \; degree$