Question

The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given by $y=(4t-2t^2)\text{m}$ and $x=\left(3t\right)\text{m}$, where $t$ is in seconds and point of projection is taken as origin. The angle of projection of projectile with vertical is

• A. ${30}^o$
• B. ${37}^o$
• C. ${45}^o$
• D. ${60}^o$

Answer 1

The correct option is B ${37}^o$

Given
$y=4t-2t^2............\left(i\right)$
We know that vertical displacement of the projectile from the ground is given by $y=u_{y}t-\frac{1}{2}{g}_p{t}^2$
On comparing equation $\left(i\right)$ with above equation, we get
$u_y=4$
$g_p=4$
Horizontal displacement of the projectile is given by
$x=u_{x}t$ and $x=3t\Rightarrow u_x=3$
$\tan \theta =\frac{u_y}{u_x}=\frac{4}{3}={53}^o$ with horizontal
$\therefore$ Angle of projection of projectile with vertical $=({90}^o-{53}^o)={37}^o$