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  • The tranjectory of projectile,projected from ground is given by <img alt="\mathrm{y=x-\frac{x^2}{10}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7By%3Dx-%5Cfrac%7Bx%5

The tranjectory of projectile,projected from ground is given by

\mathrm{y=x-\frac{x^2}{10}}, where, x & y are position along axis respectively.Find the maximum height attained by the projectile will be :

 

Option: 1

2.5 m


Option: 2

3 m


Option: 3

2 m


Option: 4

5 m


Answers (1)

best_answer

\mathrm{\begin{aligned} &\begin{aligned} \frac{d y}{d x} & =0 \Rightarrow 1-\frac{2 x}{10}=0 \\ \; \; x & =5 \mathrm{~m} \end{aligned}\\ &\begin{aligned} y_{\text {max }} & =(5)-\frac{(5)^2}{10} \\ & =5-\frac{25}{10}=2.5 \mathrm{~m} \end{aligned} \end{aligned}}

Posted by

Gautam harsolia

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