A small object of uniform density rolls up a curved surface with an initial velocity $v^{\prime}$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial position. The object is

Question

Aicharya · Accepted Answer

As $v=\sqrt{\frac{2 g h}{1+\frac{k^2}{r^2}}}$
Given $h=\frac{3 v^2}{4 g}$
$v^2=\frac{2 g h}{1+\frac{k^2}{r^2}}=\frac{2 g 3 v^2}{4 g\left(1+\frac{k^2}{r^2}\right)}=\frac{6 g v^2}{4 g\left(1+\frac{u^2}{v^2}\right)}$
$1=\frac{3}{2\left(1+\frac{k^2}{v^2}\right)}$
or $1+\frac{k^2}{r^2}=\frac{3}{2}$ or $\frac{k^2}{r^2}=\frac{3}{2}-1=\frac{1}{2}$
$k^2=\frac{1}{2} r^2$ (Equation of disc)
Hence, the object is disc.

A small object of uniform density rolls up a curved surface with an initial velocity $v^{\prime}$. It…

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