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 \(...........\,m\)
Pick an option to check your answer.
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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