Question

When a body is projected from a level ground, the ratio of it's speed in the vertical and horizontal direction is $4:3$. If the velocity of projection is $u$, the time after which the ratio of the velocities in the vertical and horizontal directions is reversed is

• A. $\frac{7u}{20g}$
• B. $\frac{35u}{10g}$
• C. $\frac{9u}{g}$
• D. $\frac{10u}{g}$

Answer 1

The correct option is A $\frac{7u}{20g}$

$\frac{V_y}{V_x}=\frac{u\sin \theta }{u\cos \theta }=\frac{4}{3}=\tan \theta \Rightarrow \sin \theta =4/5$
$\cos \theta =3/5$
$u\left(t\right)=u\cos \theta \hat{i}+(u\sin \theta -gt)\hat{j}$

$\frac{u\sin \theta -gt}{u\cos \theta }=\frac{3}{4}$

$\Rightarrow \frac{\frac{4u}{5}-gt}{\frac{3u}{5}}=\frac{3}{4}$

$\Rightarrow \frac{16u}{5}-4gt=\frac{9u}{5}\Rightarrow \frac{7u}{5}=4gt$

$\Rightarrow t=\frac{7u}{20g}$