Question

Th maximum height of a projectile is half of its range on the horizontal . If the velocity of projection is u, its range on horizontal is

Answer 1

Heyy frnd here is ur ans....

We know that at maximum height

velocity = ucos0

here ,, velocity is half of its initial velocity therefore..

u/2=ucos0

so,, theta(0) = 60

now ,

we know that ,

Range = u^2sin2theta/g

R= u^2sin2×60/g

R=u^2sin120/g

or,

R=u^2√3/2g

HOPE it helps you✅✅


Thankyou⚜⚜


Sweety@