Question

Equation of trajectory of a projectile is Y = bx - $\frac{5}{9}{x}^2$. What is the maximum height reached?

• A. 25 m
• B. 35 m
• C. 45 m
• D. 55 m

Answer 1

The correct option is C 45 m

$Y=tan\theta x-\frac{2}{2u^{2}co{s}^2\theta }{x}^2,tan\theta$ = 10
$\frac{g}{2u^{2}co{s}^2\theta }$= $\frac{5}{9}$
$\therefore$ h = 45 m