The speed of a projectile at the highest point becomes 1 2 times its initial speed. The…
Physics · JEE Main · NTA Exams — Kinematics
The speed of a projectile at the highest point becomes \(\frac{1}{\sqrt{2}}\) times its initial speed. The horizontal range of the projectile will be
- \(\frac{u^{2}}{g}\)
- \(\frac{u^{2}}{2 g}\)
- \(\frac{u^{2}}{3 g}\)
- \(\frac{u^{2}}{4 g}\)
Answer
(A) u^ 2 g
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