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

Let the particle go from (O) to (P) in time ` T. Fig. (EP) . 27.
(b) Taking vertical upwared moton of particl perpendicular to plane (OX) , from (O) to (P), we have
` y_0 =0, y=0, u)y = v_0 sin beta , a_y = (- g cos alpha ), t= T`
As ` y=y_0 +u_y t + 1/2 a_y t^2`
:. ` 0=0 + (v_0 sin beta ) T+ 1/2 (- g cos alpha ) T^@` or ` T= ( 2 v_0 sin beta )/( g cos alpha)`
(a) For horizontal range ` OP` ()= R).
taking motion of particle from (O) to (P) along (OX) direction.
Here, ` x_0 =0 ,x R , u_x =v cos beta a_x =- sin alpha ` t= T= (2 v_0 sin beta )/( g cos alpha)`
As ` x= x_0 + u_x t+ 1/2 a_xt^2`
` R= 0 + v_0 cos beta xx (2 v_0 sin beta )/( g cos alpha) + 1/2 (-g sin alpga) ((2 v_0 sin beta)/(g cos alpha))^2`
` = (2 v_0^2)/( g cos alpha ) sin cos beta - ( g sin alpha )/2 xx(4 v_0^2 sin ^2 alpha)/(g^2 cos alpha ) = (2 v_0^2)/( g cos alpga))^2`
` = (2 v_0^2 sin beta )/( g cos ^2 alpha ) [ cos beta alpha - sin alpha sin beta ] = (2 v_0^2 sin beta 0/( g cos^2 alpha ) cos [ alpha + beta ]`
(c ) Ror maximum horizontal range ltbRgt ` R= v_0^2 /( g cos ^2 alpha) [ sin 2 beta cos alpha + sin alpha +sin alpha cos 2 beta ] = )v_0^2 /(g cos ^2 alpha [ sin (2 beta + alpha ) - sin alpha ]`
` R is maximum if sin (2 beta + alpha ) = 1 or ` 2 beta + alpha = pi //2` or ` beta = ((pi //s) - alpha )/2 = (pi) /4 - ( alpha)/2`.