Question

A projectile is projected from the ground by making an angle of $60^\circ $ with the horizontal. After $1 \, s$ projectile makes an angle of $30^\circ $ with the horizontal. The maximum height attained by the projectile is (Take $g=10 \, m \, s^{- 2}$ )

• A. $\frac{\text{45}}{2} m$
• B. $\frac{\text{45}}{4} m$
• C. $\frac{\text{43}}{2} m$
• D. $\frac{\text{43}}{4} m$

Answer 1

Answer: (B)

$\text{tan 30°}=\frac{\text{v}_{0} sin 60 ^\circ - \text{gt}}{\text{v}_{0} cos 60 ^\circ }\Rightarrow \frac{1}{\sqrt{3}}=\frac{\text{v}_{0} \frac{\sqrt{3}}{2} - \text{g} \times \text{t}}{\frac{\text{V}_{0}}{2}}$
$\Rightarrow \frac{\text{v}_{0}}{2} = \frac{\text{3v}_{0}}{2} - \text{gt} \sqrt{3}$
$\Rightarrow \text{v}_{0} = \text{gt} \sqrt{3} = \text{g} \sqrt{3} \text{.}$
$\therefore H_{\text{max}}=\frac{\text{v}_{0}^{2} sin^{2} \theta }{\text{2g}}=\frac{300 \times 3/4}{\text{20}}=\frac{\text{45}}{4}m$