Question

If the horizontal range of a projectile is R and the maximum height attained by it is H, then prove that
the velocity of projection is u = √[2g(H+R2/16H)]​

Answer 1

Explanation:

maximum height can be attained only when the projectile is launched at 90°

threrfore ,

h= u^2/2g

25×2×9.8= u^2

490=u^2

7√10=u

then for range ,

the maximum range occurs at 45°

R = u^2/g

R = 490/9.8

R = 50 m