Question

The angle of projection for which the max height and the horizantal range of a projectile are equal is

Answer 1

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=−gay=−g

Let’s start with the VERTICAL MOTION:

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

vf=vi+atvf=vi+at

I like to add subscripts to help:

(vf)y=(vi)y+ayt(vf)y=(vi)y+ayt

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

∴0=Vsinθ+(−g)t∴0=Vsinθ+(−g)t

time to reach maximum height is t=Vsinθgt=Vsinθg

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

Sy=(vi)yt+12ayt2Sy=(vi)yt+12ayt2

substitute in the time to reach max height gives

Sy=VsinθVsinθg+12(−g)V2sin2θg2Sy=VsinθVsinθg+12(−g)V2sin2θg2

or

Sy=V2sin2θg−g2V2sin2θg2Sy=V2sin2θg−g2V2sin2θg2

or maximum height is Sy=V2sin2θ2gSy=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+12axt2Sx=(vi)xt+12axt2

We neglect air resistance, so ax=0ax=0

∴∴Sx=(vi)xtSx=(vi)xt

The time to reach maximum 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(vi)x=Vcosθ and t=2Vsinθgt=2Vsinθg

gives Sx=(Vcosθ)2VsinθgSx=(Vcosθ)2Vsinθg

To answer your question, we equate Sx=SySx=Sy

(Vcosθ)2Vsinθg=V2sin2θ2g(Vcosθ)2Vsinθg=V2sin2θ2g

4cosθ=sinθ4cosθ=sinθ

tanθ=4tanθ=4

θ=75.96∘