A body of mass, m is projected vertically upward with a velocity, y. At what position will the potential energy be maximum?
a
\(\frac{y}{g}\)
b
\(\frac{2y^2}{g}\)
c
\(\frac{y^2}{2g}\)
d
\(\frac{y^2}{4g}\)
Explanation
Correct Option
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Chrisbridge
1 year ago
Another way, at maximum height, potential energy is maximum
Hmax = U^2/2g = y^2/2g
