Question

A projectile is thrown with a speed of $100m/s$ making an angle of ${60}^{\circ }$ with the horizontal. Find the minimum time after which its inclination with the horizontal is ${45}^{\circ }$.

• A. $5(\sqrt{3}-1)$
• B. $5(\sqrt{2}-1)$
• C. $5(1-\sqrt{3})$
• D. $5(1-\sqrt{2})$

Answer 1

The correct option is A $5(\sqrt{3}-1)$

Given, $\theta ={60}^{\circ }$
$u$ (initial velocity) = $100m/s$.
So, velocity in horizontal direction $u_x$ (along $x$) $=100\cos {60}^{\circ }$
Velocity (initial) along vertical direction $u_y$ (along $y$) = $100\sin {60}^{\circ }$
Let after time $t$ , the inclination of the particle with the horizontal is ${45}^{\circ }$

And, at time $t$, velocity along $x=v_x$ and that along $y=v_y$
So, $\frac{v_y}{v_x}=\tan {45}^{\circ }$
$v_x=v_y$
Now, since the horizontal component of velocity remains constant,
$v_x=u_x=100\cos {60}^{\circ }$
and
$v_y=u_y-gt(\text{where},v_y=v_x=100\cos {60}^{\circ })100\cos {60}^{\circ }=100\sin {60}^{\circ }-10t100\times \frac{1}{2}=100\times \frac{\sqrt{3}}{2}-10t50=50\sqrt{3}-10tt=5(\sqrt{3}-1)s$