Discussion Forum

A projectile is given an initial velocity of $(\hat{i}+2j)$  m/s, where   $\hat{i}$  is along the ground and $\hat{j}$ is along the vertical. If g=10 m/s², the equation of its trajectory is

Answer:

Explanation:

intial velocity= (i+2j) m/s

 Magnitude  of initial velocity  $u=\sqrt{(1)^{2}+(2^{2})}=\sqrt{5}m/s$

 Equation of trajectory of projectile is

$y=x\tan\theta -\frac{gx^{2}}{2u^{2}}(1+\tan ^{2}\theta)$

                                    $\left[ \tan\theta=\frac{y}{x}=\frac{2}{1}=2\right]$

 $\therefore$       $  y=x\times2-\frac{10(x)^{2}}{2(\sqrt{5})^{2}}[1+(2)^{2}]$

$=2x-\frac{10(x^{2})}{2\times5}(1+4)$

     = 2x-5x²