Question

In horizontal level ground to ground projectile if at any instant velocity becomes perpendicular to initial velocity then what can you say about projection angle with horizontal.

• A. $\theta =45^{\circ }$
• B. $\theta \ge 45^{\circ }$
• C. $\theta \le 45^{\circ }$
• D. for any value of $\theta$ it is possible

Answer 1

Answer: (B)

Velocity at any time $\vec{v}=\vec{u}+\vec{g}t$

$\Rightarrow \vec{v}=ucos\theta \hat{i}+\left(usin\theta -gt\right)\hat{j}$

Let at any time this velocity becomes perpendicular to initial velocity. The $\vec{v}=\vec{u}=0$

Solve to get $t=\frac{u}{gsin\theta }$

Now $t$ should be less than/equal to time of flight.
So $t\le$ T.

$\frac{u}{g\sin \theta }\le \frac{2u\sin \theta }{g}$
$\Rightarrow si{n}^2\theta \ge \frac{1}{2}$

$\Rightarrow sin\theta \ge \frac{1}{\sqrt{2}}$
$\Rightarrow \theta \ge 45^{\circ }$