A particle is projected in aer an angle ` beta` to a surface which itself is inclined at an angle ` alpha` to the horixonta (Fig. 2 (EP). 26)
(a) Find an ecxpression for range on the plane surface (distanc eon the plance from the point of projection at which particle will hit the surface). (b) Time of flight. 9c ) ` beta` at which range will be maximum.

Explanation

Consider the adjacent diagram.
Mutually perpendicular x and y-axes are shown in the diagram.
Particle is projected from the point O.
Let time taken in reaching from point O to point P is T.
(b) Considering motion along vertical upward direction perpendicular to OX.
For the journey O to P.
`y=0, u_(y)=v_(0) sin beta, a_(y) = - g cos alpha, t = T`
Applying equation,
`y=u_(y)t+(1)/(2)a_(y)t^(2)`
`rArr " " 0=v_(0)sin beta T + (1)/(2) (-g cos alpha)T^(2)`
`rArr T[v_(0)sin beta -(gcos alpha)/(2)T]=0`
`rArr " " T=0,T=(2 v_(0)sin beta)/(g cos alpha)`
As T = 0, corresponding to point O
Hence, `" " T="Time of flight" =(2 v_(0)sin beta)/(g cos alpha)`
(a) Considering motion along OX,
`x=L, u_(x)=v_(0) cos beta, a_(x)=-g sin alpha`
`t=T=(2v_(0)sin beta)/(g cos alpha)`
`x=u_(x)t+(1)/(2)a_(x)t^(2)`
`rArr " " L=v_(0)cos beta T+(1)/(2)(-g sin alpha)T^(2)`
`rArr " " L=v_(0) cos betaT-(1)/(2)g sin alpha T^(2)`
`=T[v_(0)cos beta -(1)/(2) g sin alpha T]`
`=T[v_(0)cos beta-(1)/(2) g sin alpha xx (2v_(0)sin beta)/(g cos alpha)]`
`=(2v_(0)sin beta)/(g cos alpha)[v_(0)cos beta-(v_(0)sin alpha sin beta)/(g cos alpha)]`
`=(2v_(0)^(2)sin beta)/(g cos^(2) alpha)[cos beta.cos alpha-sin alpha . sin beta]`
`rArr " " L=(2v_(0)^(2)sin beta)/(g cos^(2) alpha)cos (alpha+beta)`
(c) For range (L) to be maximum.
Let, `" " Z=sin beta. cos(alpha+beta)`
`=sinbeta[cos alpha.cosbeta-sinalpha.sin beta]`
`=(1)/(2)[cosalpha.sin 2 beta - 2 sin alpha. sin^(2)beta]`
`=(1)/(2)[sin 2 beta. cos alpha - sin alpha (1-cos 2 beta)]`
`rArr " " z=(1)/(2) [sin 2 beta. cos alpha-sin alpha + sin alpha. cos 2 beta]`
`=(1)/(2)[sin 2 beta . cos alpha + cos 2 beta. sin alpha - sin alpha]`
`=(1)/(2) [sin (2beta + alpha)-sin alpha]`
For z to be maximum,
`sin (2=beta+alpha)`=maximum = 1
`rArr " " 2beta + alpha =(pi)/(2) or, beta = (pi)/(4) -(alpha)/(2)`