The velocity at the maximum height of a projectile is \( \frac{\sqrt{3}}{2} \) times its i....
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Question The velocity at the maximum height of a projectile is \( \frac{\sqrt{3}}{2} \) times its initial velocity of projection \( (u) \). Its range on the horizontal plane is
(A) \( \frac{\sqrt{3} u^{2}}{2 g} \)
(B) \( \frac{3 u^{2}}{2 g} \)
(C) \( \frac{3 u^{2}}{g} \)
(D) \( \frac{u^{2}}{2 g} \)📲PW App Link - https://bit.ly/YTAI_PWAP 🌐PW Website - https://www.pw.live