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WAEC Physics 2024 Paper

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Question : 24

Total: 50

A body of mass, m is projected vertically upward with a velocity, y. At what position will the potential energy be maximum?

$\;\frac{y}{g}$
$\;\frac{2y^2}{g}$
$\;\frac{y^2}{2g}$
$\;\frac{y^2}{4g}$

Solution:

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: $\;\frac{y^2}{2g}$

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