Question

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

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

Answer 1

The correct option is B ${37}^{\circ }$

$PE=mb-\left(\cos \theta \right)$

So $P{E}_1=MB(1-\cos )=0$

$P{E}_2=MB(1-\cos {180}^o)=2MB=2\times \frac{2.5j}{\tau }\times 0.2\tau$

or $P.E_2=1$ Joles

$were=P.E_2-P.E,=1.0=1joule$

Given that $y=4t-2t^2$ .... $1$

$x=3t\Rightarrow t=\frac{x}{3}$ putting in eqn $1$

we get

$y=4\frac{xy}{3}-2{\left(\frac{x}{3}\right)}^2$

comparing with

$y=x\tan \theta -$\dfrac {1}{2} \dfrac{9x^2}{4^2 \cos^2 \theta }$

we get $\tan \theta =\frac{4}{3}\Rightarrow \theta ={53}^o$

from horizontal

So $anglefromvertical=90-{53}^o={37}^o$