Question

the horizontal velocity of a projectile is 2/√5 times the half of the maximum speed, then the ratio of horizontal range to maximum height is​

Answer 1

Explanation:

range =2ux uy/g

max height =uu/2g

Answer 2

Answer:

Let v be velocity of a projectile at maximum height H

v = ucosθ

According to given problem, v =

2

u

∴

2

u

=ucosθ⇒cosθ=

2

1

θ=60

o