Question

A particle travels according to the equation $y=x-\frac{x^2}{2}$. Find the maximum height it achieves.

• A. $1\text{m}$
• B. $2\text{m}$
• C. $\frac{1}{2}\text{m}$
• D. $\frac{1}{4}\text{m}$

Answer 1

The correct option is C $\frac{1}{2}\text{m}$

The given equation is $y=x-\frac{x^2}{2}$
Comparing the given equation with the equation of the trajectory of a projectile.
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$, we get
$\tan \theta =1\Rightarrow \theta ={45}^{\circ }$ and
$\frac{g}{2u^2{\cos }^2\theta }=\frac{1}{2}\Rightarrow \frac{g}{u^2}=\frac{1}{2}\Rightarrow u^2=2g$
($\because \cos {45}^{\circ }=\frac{1}{\sqrt{2}}$)

$h_{\text{max}}=\frac{u^2{\sin }^2\theta }{2g}=\frac{2g{\sin }^2{45}^{\circ }}{2g}=\frac{1}{2}$
Thus, $h_{\text{max}}=\frac{1}{2}\text{m}$

Alternate:
The maximum value of a function occurs at the point given by
$\frac{dy}{dx}=0$ and $\frac{d^{2}y}{d{x}^2}<0$
Thus, we have
$\frac{dy}{dx}=1-\frac{2x}{2}=0\Rightarrow x=1$
and $\frac{d^{2}y}{d{x}^2}=-1<0$
Hence, the maximum value of the given function is
$y_{max}=1-\frac{1}{2}=\frac{1}{2}\text{m}$