The trajectory of projectile, projected from the ground is given by $y=x-\frac{{x}^{2}}{20}$. Where x and y are measured in meter. The maximum height attained by the projectile will be:

Question

YESPYQ · Accepted Answer

Correct answer: a) $5\mathrm{m}$. $y=x-\frac{{x}^{2}}{20}$ For maximum height, $\frac{dy}{dx}=0\Rightarrow \frac{d}{dx}\left(x-\frac{{x}^{2}}{20}\right)=0\ 1-\frac{2x}{20}=0\ x=10$ So, ${y}_{\max }=10-\frac{100}{20}=5m$

The trajectory of projectile, projected from the ground is given by \(y=x-\frac{{x}^{2}}{20}\). Where x and y are measur…

Practice more NEET Physics PYQs