Question

A projectile is thrown with a velocity of $20 \, m / s$ , at an angle of $60^\circ $ with the horizontal. After how much time the velocity vector will make an angle of $45^\circ $ with the horizontal

• A. $\sqrt{3} \, s$
• B. $\frac{1}{\sqrt{3}} \, s$
• C. $\left(\sqrt{3} + 1\right) \, s$
• D. $\left(\sqrt{3} - 1\right) \, s$

Answer 1

Answer: (D)

As the horizontal component is the same for a projectile
Let projectile velocity is $v$ when particle makes angle $45^\circ $ with horizontal,
hence
$v=ucos \theta sec ⁡ \phi=20cos ⁡ 60^\circ sec ⁡ 45^\circ $
$v=20.\frac{1}{2}.\sqrt{2}=10\sqrt{2} \, \, m/s$
$t=\frac{v_{y} - u_{y}}{g}=\frac{10 \sqrt{3} - 10}{10}=\left(\sqrt{3} - 1\right) \, s$