Question

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

• A. $y=x-5x^2$
• B. $y=2x-5x^2$
• C. $4y=2x-5x^2$
• D. $4y=2x-25x^2$

Answer 1

Answer: (B)

Initial 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)\bigg[tan \theta=\frac{y}{x}=\frac{2}{1}=2\bigg]$
$\therefore \, \, \, y=x \times 2-\frac{10(x)^2}{2(\sqrt 5)^2}[1+(2)^2]$
$ \, \, \, \, \, \, \, \, \, \, \, \, =2x-\frac{10(x^2)}{2 \times 5}(1+4)$
$ \, \, \, \, \, \, \, \, \, \, \, \, 2x-5x^2$