JEE Mains · Physics · STD 11 - 3.2 motion in plane
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 \(15\)
- D \(16\)
Answer & Solution
Correct Answer
(D) \(16\)
Step-by-step Solution
Detailed explanation
\(\mathrm{H}_{\max }=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\) \(\frac{\mathrm{H}_{1 \max }}{\mathrm{H}_{2 \max }}=\frac{\mathrm{u}_1^2}{\mathrm{u}_2^2}\) \(\frac{64}{\mathrm{H}_{2 \max }}=\frac{\mathrm{u}^2}{(\mathrm{u} / 2)^2}\) \(\mathrm{H}_{2 \max }=16 \mathrm{~m}\)
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