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JEE Mains · Physics · STD 11 - Motion in plane
  • JEE Mains 2023
  • Medium
  • Numerical

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 \(...........\,m\)

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Answer & Solution

Correct Answer

(A) \(5\)

Step-by-step Solution

Detailed explanation

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

For maximum height,

\(\frac{ dy }{ dx }=0 \Rightarrow 1-\frac{2 x }{20}=0\)

\(x=10\)

So, \(y _{\max }=10-\frac{100}{20}=5\,m\)

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