A body of mass, m is projected vertically upward with a velocity, y. At what position will the potential energy be maximum?
- \(\frac{y}{g}\)
- \(\frac{2y^2}{g}\)
-
\(\frac{y^2}{2g}\)
- \(\frac{y^2}{4g}\)
The correct answer is: C
Explanation
At the maximum height, all the kinetic energy will have been converted into potential energy:
mgh = \(\frac{1}{2}\)mv\(^2\)
v = y
gh = \(\frac{1}{2}\)y\(^2\)
h = \(\frac{y^2}{2g}\)
Thus, the potential energy will be maximum at the height given by: \(\frac{y^2}{2g}\)