MCQ
The maximum height reached by a projectile is $64 \mathrm{~m}$. If the initial velocity is halved, the new maximum height of the projectile is_________.$\mathrm{m}$.
- A$11$
- B$14$
- C$15$
- ✓$16$
✓
Answer
Correct option: D.
$16$
d
$\mathrm{H}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}$
$\mathrm{H}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}$
$\frac{\mathrm{H}_{1 \max }}{\mathrm{H}_{2 \max }}=\frac{\mathrm{u}_1^2}{\mathrm{u}_2^2}$
$\frac{64}{\mathrm{H}_{2 \max }}=\frac{\mathrm{u}^2}{(\mathrm{u} / 2)^2}$
$\mathrm{H}_{2 \max }=16 \mathrm{~m}$
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