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 $\vec{v}=\vec{u}+\vec{g}t$
$\Rightarrow \vec{v}=u \,cos \,\theta\hat{i} +\left(u \,sin\, \theta-gt\right)\hat{j}$
Let at any time this velocity becomes perpendicular to initial velocity. The $\vec{v}=\vec{u}=0$
Solve to get $t=\frac{u}{g\,\,sin\,\, \theta}$
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 sin^{2} \theta \ge\frac{1}{2}$
$\Rightarrow sin\, \theta \ge\frac{1}{\sqrt{2}} $
$\Rightarrow \theta \ge 45^{\circ}$