Question

The maximum height attained by a projectile is found to be equal to $0.433$ of horizontal range. The angle of projection of this projectile is

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${75}^0$

Answer 1

Answer: (C)

${60}^0$

According to question,

$H=0.433R$

$\frac{u^{2}si{n}^2\theta }{2g}=0.433\frac{u^{2}sin2\theta }{g}$

$si{n}^2\theta =2\times 0.433\times 2sin\theta .cos\theta$

$tan\theta =4\times 0.433=1.732$

$\theta =ta{n}^{-1}\left(1.732\right)={59.99}^0$

$\theta \approx {60}^0$

The correct option is C.