Question

The velocity at the maximum height of a projectile is half its initial velocity of projection v0. What is the horizontal range of the projectile?

Answer 1

Answer:

√3u²/2g

Explanation:

Velocity at maximum height = half of initial velocity.

Velocity at maximum height = horizontal component.

v(X)=u/2

v(y)=√(u²-u²/4)

=√3u/2

v(y)/v(x)=√3

tanø=√3

ø=60°

R=2u²sinøcosø/g

=√3u²/2g