Question

वेलोसिटी ऑफ द प्रोजेक्टाइल अट एनी पॉइंट ऑन द प्रोजेक्ट ट्री​

Answer 1

The velocity at the maximum height of a projectile is half of its velocity of projection

u cos theta = (u)/(2) rArr cos theta = (1)/(2) rArr theta = 60^(@)

R = (u^(2))/(d) sin 2 theta = (u^(2))/(g) sin 120^(@) = (u^(2))/(g) sin 60^(@)

= (sqrt(3) u^(2))/(2g)

I hope it will help you