Question

the speed of projectile at its maximum height is 1/2 times its initial speed , u.. its range on horizontal plane is?​

Answer 1

speed at the maximum height is only going to be horizontal component which is

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