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  • The trajectory of projectile, projected from the ground is given by <img alt="\mathrm{y}=\mathrm{x}-\frac{\mathrm{x}^{2}}{20}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7By%7D%

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

Option: 1

10 \mathrm{~m}


Option: 2

200 \mathrm{~m}


Option: 3

10 \sqrt{2} \mathrm{~m}


Option: 4

5 \mathrm{~m}


Answers (1)

best_answer

y=x-\frac{x^{2}}{20}

\left(\frac{d y}{d x}\right)=1-\frac{x}{10} \text { for } y_{\text {max }} ; \frac{d y}{d x}=0

x=10

y_{\max }=10-\frac{100}{20}=5 \mathrm{~m}

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