Question

A body is thrown with the velocity $v_0$ at an angle of $\theta$ to the horizon. Determine $v_0$ in $m{s}^{-1}$ if the maximum height attained by the body is $5m$ and at the highest point of its trajectory the radius of curvature is $r=3m$. Neglect air resistance. [Use $\sqrt{80}$ as $9$].

Answer 1

Radius of curvature at the highest point
$r=\frac{v}{g}\Rightarrow r=\frac{(v_0\cos \theta {)}^2}{g}$
$v_0\cos \theta =\sqrt{gr}....\left(i\right)$
Maximum height $h=\frac{v_0^2{\sin }^2\theta }{g}\Rightarrow v_0\sin \theta =\sqrt{gh}....\left(ii\right)$
By using (i) and (ii), we get
$v_0^2{\sin }^2\theta +v_0^2{\cos }^2\theta =gh+gr=g(h+r)$
$\Rightarrow v_0=\sqrt{g(h+r)}=\sqrt{10(5+3)}=9m{s}^{-1}$.