MCQ
The maximum horizontal range of a projectile is $160\, m$. When the projectile is thrown with the same speed at an elevation of $30^o$ from the horizontal, it will reach to the maximum height of ......... $m$
- ✓$20$
- B$40$
- C$80 $
- D$160$
✓
Answer
Correct option: A.
$20$
a
$\mathrm{R}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}$ so that $\mathrm{R}_{\max }=\frac{\mathrm{u}^{2}}{\mathrm{g}}=160 \mathrm{m}$
$\mathrm{R}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}$ so that $\mathrm{R}_{\max }=\frac{\mathrm{u}^{2}}{\mathrm{g}}=160 \mathrm{m}$
$\mathrm{H}_{\max }=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}}$
$=\frac{1}{2} \times 160 \times \sin ^{2} 30^{\circ}=20$
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