# Q. A projectile is thrown from the base of an inclined plane at an angle 45° with the plane.At what angle may it hit the plane again? A) 30° B) 45° C) 60° D) None of these

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### 2 Answers

Let the angle of inclination of the plane be `alpha` . Further assume the inclined plane to be the x-axis and its perpendicular plane to be the y-axis.

The equations of motion of the projectile :

`v_x=ucos45^o-gsinalpha*t`

`v_y=usin45^o-gcosalpha*t`

`S_x=ucos45^o*t-1/2gsinalpha*t^2`

`S_y=usin45^o*t-1/2gcosalpha*t^2`

At the point of landing, `S_y=0`

So, `usin45^o*t-1/2gcosalpha*t^2=0`

`rArr t= (2usin45^o)/(gcosalpha)`

The landing angle l is given by:

`tanl=-v_y/v_x=-(usin45^o-gcosalpha*t )/(ucos45^o-gsinalpha*t)`

Put the value of t for landing,

`tanl= (u/sqrt2-gcosalpha*2u/sqrt2/(gcosalpha))/(u/sqrt2-gsinalpha*(2u/sqrt2)/(gcosalpha))`

`= (u(1/sqrt2-sqrt2))/(u(1/sqrt2-sqrt2*tanalpha))`

`= (1/sqrt2-sqrt2)/(1/sqrt2-sqrt2*tanalpha)`

Clearly, the landing angle, l would depend upon the angle of inclination, `alpha` and should vary with variation of `alpha` .

So, the correct answer is option D.

Here is a pictorial representation of the motion: