Answer to Question #108078 in Mechanics | Relativity for Maro

Question #108078
A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of a cliff 75.0 m high.

(a).How much later does it reach the bottom of the cliff?

(b).What is its speed just before hitting?

(c).What total distance did it travel?
1
Expert's answer
2020-04-06T09:08:28-0400

a)


"y=0=h+vt-0.5gt^2"

"0=75+15t-0.5(9.8)t^2"

"t=5.73\\ s"

b)


"V^2=v^2+2gh"

"V^2=15^2+2(9.8)(75)"

"V=41.2\\frac{m}{s}"

c)


"s=h+2\\frac{v^2}{2g}=h+\\frac{v^2}{g}"

"s=75+\\frac{15^2}{9.8}=98.0\\ m"


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