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

Options

A. `(2u^(2)sin alpha cos(alpha + beta))/(g cos^(2) alpha)`
B. `(2u^(2) sin beta cos(alpha + beta))/(g cos^(2) beta)`
C. `(2u^(2) sin beta cos (alpha + beta))/(g cos^(2) alpha)`
D. `(2 u^(2) sin alpha cos (alpha + beta))/(g cos^(2) beta)`

Answer

C

Explanation

Take the x-axis along the incline and y-axis perpendicular to the plane.
`:.u_(x)ucosbeta`
`u_(y)=usinbeta`
`a_(x)=-gsinalpha`
`a_(y)=-gcosalpha`
When the particle lands at P its y coordinate becomes zero.
`:.0=u_(y)t+(1)/(2)a_(y)t^(2)`
`0=usinbetat-(1)/(2)gcosalphat^(2)ort=(2usinbeta)/(gcosalpha)" "......(i)`
For motion along inclined plane, `x=u_(x)t+(1)/(2)a_(x)t^(2)`
`:.L=ucosbetat-(1)/(2)gsinalphat^(2)`
Substituting the value of t from eq. (i), we get
`L=ucosbeta((2usinbeta)/(gcosalpha))-(1)/(2)gsinalpha((2usinbeta)/(gcosalpha))^(2)`
`=(2u^(2)sinbetacosbeta)/(gcosalpha)-(2u^(2)sinalphasin^(2)beta)/(gcos^(2)alpha)`
`=(2u^(2)sinbetacosbeta)/(gcosalpha)-(2u^(2)sinalphasin^(2)beta)/(gcos^(2)alpha)`
`=(2u^(2)sinbeta)/(gcos^(2)alpha)[cosbetacosalpha-sinalphasinbeta]=(2u^(2)sinbetacos(alpha+beta))/(gcos^(2)alpha)`