A projectile is given an initial velocity of $(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \mathrm{ms}^{-1}$. The equation of its path is $\left(g=10 \mathrm{~ms}^{-2}\right)$

• Correct answer
  A

  $y=2 x-5 x^2$
• B

  $y=x-5 x^2$
• C

  $4 y=2 x-5 x^2$
• D

  $y=2 x-25 x^2$

Solution

Velocity of particle is $(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}) \mathrm{ms}^{-1}$ initially. So, $\quad u_x=1 \mathrm{~ms}^{-1}$ and $u_y=2 \mathrm{~ms}^{-1}$ Also, $\quad a_x=0$ and $a_y=-10 \mathrm{~ms}^{-2}$ In time $t$, Horizontal distance covered by projectile is
And vertical distance covered by projectile is $$ y=u_y t+\frac{1}{2} a_y t^2 $$
Substituting the value of $t$ from Eq (i) in Eq (ii), we get $$ y=2 x-5 x^2 $$

Hence, the correct answer is option: 'A'.