A particle projected from ground moves at an 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]

The correct option is B ${\tan }^{-1}\left(2\right)$
Let $u$ be the initial velocity of the projectile and suppose it makes an angle $\alpha$ with the horizontal.
Let $v$ be the velocity of the projectile when it makes an angle of ${45}^{\circ }$ with the horizontal.
Vertical component of velocity at this point is given by
$v_y=u_y-g\left(1\right)$$......\left(i\right)$
(by first equation of motion)
Horizontal component of velocity at this point is given by
$v_x=u_x$ (Acceleration is zero along horizontal component) $......\left(ii\right)$
Dividing $\left(i\right)$ from $\left(ii\right),$ we get
$\frac{v_y}{v_x}=\tan {45}^{\circ }=\frac{u_y-g}{u_x}...\left(iii\right)$
Let $T$ be the time of flight of the projectile
According to question.
$\frac{T}{2}=2s$
(Particle reaches maximum height in $2s$)
$\Rightarrow \frac{2u\sin \alpha }{g}=4$
$\Rightarrow u\sin \alpha =u_y=2g...\left(iv\right)$
Put $\left(iv\right)$ in $\left(iii\right)$, we get
$1=\frac{2g-g}{u\cos \alpha }$
$\Rightarrow u\cos \alpha =g......\left(v\right)$
Dividing $\left(iv\right)$ from $\left(v\right)$, we get
$\tan \alpha =2$
$\therefore \alpha ={\tan }^{-1}2$
Hence, the correct answer is option (b)