4.31. A particle is projected in the air at an angle β to a surface which itself is inclined at an angle α to the horizontal.

a) find an expression of range on the plane surface

b) time of flight

c) β at which range will be maximum


Hint: Time taken to travel the distances along the inclined plane and perpendicular to the inclined plane will be the same.
Step 1: Consider the adjacent diagram.
solution

Mutually perpendicular x and y-axes are shown in the diagram.
The particle is projected from point O.
Let the time taken in reaching from point O to point P is T.

(b) Step 2: Considering motion along a vertically upward direction perpendicular to OX.

For the journey O to P.
y=0,uy=v0sinβ,ay=gcosαit=T
Applying equation,

y=uyt+12ayt20=v0sinβT+12(gcosα)T2Tv0sinβgcosα2T=0

T=0,T=2v0sinβgcosα As T=0, corresponding to point O Hence, T= Time of flight =2v0sinβgcosα

(a) Step 3: Considering motion along OX.

x=L1, ux=v0cosβ, ax=gsinαt = T = 2v0sinβgcosαx = uxt+12axt2

 L = v0cosβT + 12(gsinα)T2 L = v0cosβT  12gsinαT2= T[v0cosβ12gsinαT]= T[v0cosβ12gsinα×2v0sinβgcosα]= 2v0sinβgcosα[v0cosβ  v0sinαsinβcosα]=2v02sinβgcos2α[cosβcosαsinαsinβ] L = 2v02sinβgcos2αcos(α+β)

(c) Step 4: For range (L) to be maximum, sinβcos(α+β) should be maximum

Let, Z=sinβcos(α+β)=sinβcosαcosβsinαsinβ=12cosαsin2β2sinαsin2β=12[sin2βcosαsinα(1cos2β)] z=12[sin2βcosαsinα+sinαcos2β]=12[sin2βcosα+cos2βsinαsinα]=12[sin(2β+α)sinα]

Step 5: For z to be maximum

sin(2=β+α)= maximum =1 2β+α=π2 or, β=π4α2