Question

A 1.5 kilogram projectile launched at some angle leaves the ground with kinetic energy totally 250 joules at the highest height, its kinetic energy is 105 joules. To what vertical height did the projectile rise?

Answer 1

Given:

$m=1.5\:\rm kg$

$E_{k1}=250\:\rm J$

$E_{p1}=0\:\rm J$

$E_{k2}=105\:\rm J$

The law of conservation of energy says

$$E_{k1}+E_{p1}=E_{k2}+E_{p2}$$

or

$$E_{k1}=E_{k2}+mgh_2$$

Hence, the maximum height

$$h_2=\frac{E_{k1}-E_{k2}}{mg}=\frac{250-105}{1.5*9.8}=9.9\:\rm m$$