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  4. A stone is projected vertically up with a velocity U, reaches upto a maximum height h. When it is at a height of (3 h/4) from the ground, the ratio of KE and PE at that point is: (.PE=0 at the point of projection)

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Q. A stone is projected vertically up with a velocity $U,$ reaches upto a maximum height h. When it is at a height of $\frac{3 h}{4}$ from the ground, the ratio of $KE$ and $PE$ at that point is : $\left(\right.PE=0$ at the point of projection)

NTA AbhyasNTA Abhyas 2020

Solution:

$\frac{1}{2}mu^{2}=mgh$
at a height $\frac{3 h}{4}P.E.$ $=mg\left(\frac{3 h}{4}\right)$
$K.E=mg\left(\frac{h}{4}\right)$
$KE:$ pot. energy
$1:3$