(1)using the coordinate system as described,the path of the ball is determined byy=Vsinα·t-1/2·g·t^2x=Vcosα·t(2)rotate the original coordinate system by an angle β to establish a new one, which preserves the orgin but uses the inclined plane as the X axisbased on results from question 1, we have the new path of the ball as:y'=Vsin(α-β)·t-1/2·gcosβ·t^2x'=Vcos(α-β)·t-1/2·gsinβ·t^2when the ball lands on the plane, y'=0hencet=2Vsin(α-β)/(gcosβ)(3)further to the results of question 2, when the ball lands on the inclined plane,x'=Vcos(α-β)·2Vsin(α-β)/(gcosβ)-1/2·gsinβ·[2Vsin(α-β)/(gcosβ)]^2=2·V^2·[sin(α-β)cos(α-β)-sin(α-β)·sin(α-β)tgβ]/(gcosβ)x' reaches its maximum when its derivative equals to zeroie. derivative of sin(α-β)cos(α-β)-sin(α-β)·sin(α-β)tgβ equals to zero.Hence cos[2(α-β)]-sin[2(α-β)]tgβ=0 ctg[2(α-β)]=tgβ α-β=(90-β)/2That is to say, maximum range is achieved when α bisects the angle between the plane and the verticalBy plugging α-β=(90-β)/2 into the x' formulation,you get the maximum.
小妮子--Amy
