Question

A particle is projected at an angle of ${30}^o$ to the horizontal and $2$ seconds later is moving in the direction ${\tan }^{-1}1/4$ to the horizontal for the first time. Find its initial speed. $[g=10{m/s}^2]$.

Answer 1

Equation at time $t=0\sec$

$v_y=u_y$

$v_x=u_x$

$\tan \theta =\frac{u_y}{u_x}$

Equation at time $t=2\sec$

$v_y=u_y-g\left(2\sec \right)$

$v_x=u_x$

$\tan {\theta }^{\prime }=\frac{u_y-g\left(2\sec \right)}{u_x}$

At time $t=0\sec$$\tan \theta =\frac{1}{\sqrt{3}}$

At time $t=2\sec$$\tan {\theta }^{\prime }=\frac{1}{4}$

Initial velocity will be

$=\sqrt{u_x^2+u_y^2}$

Solving for $u_x$ and $u_y$ we get,

$u_x=\frac{8\sqrt{3g}}{4-\sqrt{3}}m/s$ and $u_x=\frac{8g}{4-\sqrt{3}}m/s$

And initial velocity will be,

$=\frac{16g}{4-\sqrt{3}}m/s$

$=\frac{160}{4-\sqrt{3}}m/s$