Question

The trajectory of a projectile is given by the equation $y=x-x^2$. What is the value of the initial velocity and the angle of projection?

• A. $\sqrt{10}\text{m/s},{30}^{\circ }$
• B. $10\text{m/s},{45}^{\circ }$
• C. $\sqrt{10}\text{m/s},{45}^{\circ }$
• D. $10\text{m/s},{30}^{\circ }$

Answer 1

The correct option is C $\sqrt{10}\text{m/s},{45}^{\circ }$

The equation of a projectile is given by $y=x\tan \theta -\frac{1}{2}\frac{g{x}^2}{u^2{\cos }^2\theta }$.
Comparing the standard equation with the given equation $y=x-x^2$, we get
$\tan \theta =1$
$\Rightarrow \theta ={\tan }^{-1}\left(1\right)={45}^{\circ }$
Also, $\frac{1}{2}\frac{g}{u^2{\cos }^2\theta }=1$
$\Rightarrow \frac{1}{2}\frac{10}{u^2{\cos }^2{45}^{\circ }}=1$
$\Rightarrow \frac{1}{2}\frac{10\times 2}{u^2}=1$
$\Rightarrow u=\sqrt{10}\text{m/s}$