JEE Mains · Physics · STD 11 - 3.2 motion in plane
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 \(...........\,m\)
- A \(5\)
- B \(10 \sqrt{2}\)
- C \(200\)
- D \(10\)
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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