First slide

Projectile motion

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?

Difficult

A

$θ = 45^{\circ}$
B

$θ ⩾ 45^{\circ}$
C

$θ < 45^{\circ}$
D

for any value of $θ$ it is Possible

Solution

Velocity at any time, $\vec{v} = \vec{u} + \vec{g} t$
$\Rightarrow \vec{v} = u \cos θ \hat{i} + (u \sin θ - g t) \hat{j}$
Let at any time, this velocity becomes perpendicular to initial velocity. Then $\vec{v} \cdot \vec{u} = 0$
Solve to get $t = \frac{u}{g \sin θ}$
Now t should be less than/equal to time of flight (T).
$So t \leq T$
$\begin{array}{l} \frac{u}{g \sin θ} \leq \frac{2 u \sin θ}{g} \Rightarrow \sin^{2} θ \geq \frac{1}{2} \\ \Rightarrow \sin θ \geq \frac{1}{\sqrt{2}} \Rightarrow θ \geq 45^{\circ} \end{array}$

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