MCQ
The speed at the maximum height of a projectile is half of its initial speed $u$. The horizontal range of projectile is
- A$\frac{3 u^2}{2 g}$
- B$\frac{\sqrt{3} u^2}{2 g}$
- C$\frac{u^2}{2 g}$
- D$\frac{2 u^2}{g}$
✓
Answer
(B)Projectile velocity at maximum height
$\begin{aligned} v & =u \cos \theta \\ \frac{u}{2} & =u \cos \theta \\ \frac{1}{2} & =\cos \theta \\ \theta & =60^{\circ}\end{aligned}$
Horizontal range
$\begin{array}{l} R =\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \sin 120^{\circ}}{g} \\ R =\frac{\sqrt{3} u^2}{2 g} \quad \because \sin 120^{\circ}=\frac{\sqrt{3}}{2}\end{array}$
Hence, the correct option is (b).
$\begin{aligned} v & =u \cos \theta \\ \frac{u}{2} & =u \cos \theta \\ \frac{1}{2} & =\cos \theta \\ \theta & =60^{\circ}\end{aligned}$
Horizontal range
$\begin{array}{l} R =\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \sin 120^{\circ}}{g} \\ R =\frac{\sqrt{3} u^2}{2 g} \quad \because \sin 120^{\circ}=\frac{\sqrt{3}}{2}\end{array}$
Hence, the correct option is (b).
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