Question

The parabolic path of projectile is Y= $\frac{x}{\sqrt{3}}-\frac{x^2}{60}$ its maximum height is

• A. 5 m
• B. 10 m
• C. 15 m
• D. 20 m

Answer 1

The correct option is A 5 m

Comparing with the standard equation of trajectory of a projectile
$tan\theta =\frac{1}{\sqrt{3}};\theta ={30}^{0}R=\frac{a}{b}=\frac{60}{\sqrt{3}}H=\frac{Rtan\theta }{4}=\frac{\left(\frac{60}{\sqrt{3})}\right(\frac{1}{\sqrt{3}})}{4}=5m$