Answer to Question #189674 in Physics for Ellen

Question

When an object is thrown vertically upward it raises to maximum height of 30m. How high is the object when it has lost one third of its kinetic energy? What is the speed of the object at this point?

Answer 1

$mgH=mv^2/2$

$\frac{1}{3}KE=\frac{1}{3}\cdot\frac{mv^2}{2}=\frac{1}{3}\cdot mgH$

$\frac{1}{3}\cdot mgH=mgh\to h=H/3=30/3=10\ (m)$ . Answer

$mgH-\frac{1}{3}\cdot mgH=\frac{2}{3}\cdot mgH$

$\frac{2}{3}\cdot mgH=\frac{mv^2}{2}\to v=\sqrt{\frac{4gH}{3}}=\sqrt{\frac{4\cdot9.81\cdot 30}{3}}=19.8\ (m/s)$ . Answer