Question

A projectile is given an initial velocity of $\hat{i}+2\hat{j}$. The cartesian equation of its path is (Take g = 10 m s^-2)

• A. y = x - 5x²
• B. y = 2x - 5x²
• C. y = 2x - 15x²
• D. y = 2x - 25x²

Answer 1

Answer: (B)

Given, $\vec{u}=\hat{i}+2\hat{j}=u_x\hat{i}+u_y\hat{j}$
Then $u_x=1=ucos\theta$
and $u_y=2=usin\theta$
$\therefore tan\theta =\frac{usin\theta }{ucos\theta }=\frac{2}{1}=2$
The equation of trajectory of a projectile motion is
$y=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }=xtan\theta -\frac{g{x}^2}{2{\left(ucos\theta \right)}^2}$
$\therefore y=x\times 2-\frac{10\times x^2}{2{\left(1\right)}^2}=2x-5x^2$