Question

A ball is projected horizontally from the top of a tower with a velocity $v_0$. It will be moving at an angle of ${60}^o$ with the horizontal after time -

• A. $\frac{v_0}{\sqrt{3}g}$
• B. $\frac{\sqrt{3}{v}_0}{g}$
• C. $\frac{v_0}{g}$
• D. $\frac{v_0}{2g}$

Answer 1

The correct option is B $\frac{\sqrt{3}{v}_0}{g}$

$\begin{array}{c}V\cos {60}^{\circ }=V_0\\ \Rightarrow V=2V_o\\ \therefore Vy=V\sin {60}^{\circ }=2V_{\circ }.\frac{\sqrt{3}}{2}=\sqrt{3}{V}_o\\ \Rightarrow Vg=gt\\ \Rightarrow t=\frac{\sqrt{3}{V}_o}{g}\end{array}$

Hence, the option $B$ is the correct answer.