Question

A particle projected from ground moves at angle ${45}^{\circ }$ with horizontal one second after projection and speed is minimum, two seconds after the projection. The angle of projection of particle is [Neglect the effect of air resistance].

• A. ${\tan }^{-1}\left(3\right)$
• B. ${\tan }^{-1}\left(2\right)$
• C. ${\tan }^{-1}\left(\sqrt{2}\right)$
• D. ${\tan }^{-1}\left(4\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(2\right)$

We know that

$2=t=\frac{4sin\theta }{g}$

$\frac{4sin\theta }{g}=2$

$4sin\theta =2g=20$

Now,

$t=1$

$V_x=4cos\theta$

$V_g=4cos\theta -gt$

$\therefore tan{45}^{\circ }=\frac{V_g}{V_x}$

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

$=4cos\theta =4sin\theta -gt$

and $4sin\theta =2g$

$\therefore 4cos\theta =2g-g\times 1=g$

Now

$\frac{4sin\theta =2g}{4cos\theta }=g$

$tan\theta =2$

$\theta =ta{n}^{-}\left(02\right)$