Question

A projectile is projected from the level ground in vertical x-y plane (vertical is y axis). The point of projection is considered as the origin. For which of the following initial velocities of projection u and angle of projection $\theta$ with the horizontal will the projectile pass through the point (4, 2) (all in m) (Neglect air friction)

• A. $u=4\sqrt{5}\frac{m}{s},\theta ={45}^{\circ }$
• B. $u=4\sqrt{5}\frac{m}{s},\theta =ta{n}^{-1}\left(3\right)$
• C. $u=2\sqrt{5}\frac{m}{s},\theta ={45}^{\circ }$
• D. $u=8\sqrt{5}\frac{m}{s},\theta =ta{n}^{-1}\left(3\right)$

Answer 1

Answer: (A), (B)

The correct options are
A $u=4\sqrt{5}\frac{m}{s},\theta ={45}^{\circ }$
B $u=4\sqrt{5}\frac{m}{s},\theta =ta{n}^{-1}\left(3\right)$

$x=u_{x}t=\left(ucos\theta \right)t$
$\Rightarrow t=\frac{x}{ucos\theta }$
Also, $y=u_{y}t-\frac{1}{2}g{t}^2$
$\Rightarrow y=\left(usin\theta \right)\frac{x}{ucos\theta }-\frac{1}{2}g{\left(\frac{x}{ucos\theta }\right)}^2$
$\Rightarrow y=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }$
$\left(A\right)2=4\times 1-\frac{1}{2}\frac{10\times 4^2}{16\times 5\times \frac{1}{2}}$
$\left(B\right)2=4\times 3-\frac{1}{2}\frac{10\times 4^2}{16\times 5\times \frac{1}{10}}$
$\left(C\right)2=4\times 1-\frac{1}{2}\frac{10\times 4^2}{4\times 5\times \frac{1}{2}}$
$\left(D\right)2=4\times 3-\frac{1}{2}\frac{10\times 4^2}{64\times 5\times \frac{1}{10}}$