Answer to Question #199779 in Mechanics | Relativity for Rayquel

Question

A ball has a mass of 2kg.It is dropped from a cliff and strikes the ground below at 10m/s.

1) Which is the kinetic energy as it is about to strike the ground?Ek=mv²

2) What was its potential energy before it was dropped? Ep=mgh

3) Determine the height from which was it was dropped? E=mgh

Answer 1

Solution.

$m=2kg;$

$v=10m/s;$

$1) E_k=\dfrac{mv^2}{2};$

$E_k=\dfrac{2kg\sdot100m^2/s^2}{2}=100J;$

$2)E_p=E_k=100J;$

$3)E_p=mgh\implies h=\dfrac{E_p}{mg};$

$h=\dfrac{100J}{2kg\sdot9.8m/s^2}=5.1m;$

Answer:$1)100J;$

$2)100J;$

$3)5.1m.$