Question

the velocity at the maximum height of a projectile is 0.8 times of its initial velocity u its range on the horizontal plane is​

Answer 1

I jope this mightt help you

Answer 2

Answer:

Range on the horizontal plane is $\frac{0.95u^{2} }{g}$

Explanation:

Velocity at max. height = 0.8 times of initial velocity

Velocity at max. height = horizontal component

$v_{x}$ = 0.8u

ucosθ = 0.8u

i.e cosθ = 0.8

θ = 36.86

Range of projectile = $u^{2}$sin2θ/g

R = $u^{2}$sin (73.72)/g

R = 0.95$u^{2}$/g