Question:medium
A body is thrown up with a speed \( u \), at an angle of projection \( \theta \). If the speed of the projectile becomes \( \frac{u}{\sqrt{2}} \) on reaching the maximum height, the maximum vertical height attained by the projectile is
A body is thrown up with a speed \( u \), at an angle of projection \( \theta \). If the speed of the projectile becomes \( \frac{u}{\sqrt{2}} \) on reaching the maximum height, the maximum vertical height attained by the projectile is
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At highest point, only horizontal velocity remains.
Updated On: May 2, 2026
- $\frac{u^2}{4g}$
- $\frac{u^2}{3g}$
- $\frac{u^2}{2g}$
- $\frac{u^2}{g}$
- $\frac{2u^2}{g}$
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The Correct Option is A
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