Question

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

• 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)

$\text{tan 30°}=\frac{{\text{v}}_{0}sin60^{\circ }-\text{gt}}{{\text{v}}_{0}cos60^{\circ }}\Rightarrow \frac{1}{\sqrt{3}}=\frac{{\text{v}}_0\frac{\sqrt{3}}{2}-\text{g}\times \text{t}}{\frac{{\text{V}}_0}{2}}$
$\Rightarrow \frac{{\text{v}}_0}{2}=\frac{{\text{3v}}_0}{2}-\text{gt}\sqrt{3}$
$\Rightarrow {\text{v}}_0=\text{gt}\sqrt{3}=\text{g}\sqrt{3}\text{.}$
$\therefore H_{\text{max}}=\frac{{\text{v}}_0^{2}si{n}^2\theta }{\text{2g}}=\frac{300\times 3/4}{\text{20}}=\frac{\text{45}}{4}m$