Question

Velocity at max height of projectile is 1/2 if it's initial velocity u its range on horizontal plane is

Answer 1

Your question states that at the maximum height it only has horizontal velocity = V/2

V2=Vcosθ

or

cosθ=12

or

θ=60∘

Below is my derivation of the familiar “range” equation:

Always watch your sign convention when using Newton’s equations of motion. I always assume positive = up and negative = down, so the acceleration of gravity is *always* negative:

ay=−g

Let’s start with the VERTICAL MOTION:

Use this equation of motion to find the time to reach the maximum height:

vf=vi+at

I like to add subscripts to help:

(vf)y=(vi)y+ayt

Since (Vi)y=Vsinθ and at maximum height, (vf)y=0

∴0=Vsinθ+(−g)t

time to reach maximum height is t=Vsinθg

To find the maximum height we write another familiar equation of motion using subscripts:

Sy=(vi)yt+12ayt2

substitute in the time to reach max height gives

Sy=VsinθVsinθg+12(−g)V2sin2θg2

or

Sy=V2sin2θg−g2V2sin2θg2

or maximum height is Sy=V2sin2θ2g

Now consider the HORIZONTAL MOTION:

Writing the same equation of motion except using subscripts for motion in the x-direction gives:

Sx=(vi)xt+12axt2

We neglect air resistance, so ax=0

∴Sx=(vi)xt

The time to reach the horizontal range will be twice the time to reach maximum height because the time to go up is exactly the same as the time to come back down.

Substituting in (vi)x=Vcosθ and t=2Vsinθg

gives Sx=(Vcosθ)2Vsinθg=2V2gsinθcosθ

but using the trig identity sin(2θ)=2sinθcosθ

gives

Range=Sx=V2gsin(2θ)

Answer to your question:

Substituting θ=60∘into my range equation gives

Range=V2gsin(2(60))

or

Range=0.866V2g