Question

With what angle should a particle be projected on an inclined plane so that it lands perpendicular to the plane

Answer 1

Answer:

Consider a particle projected with an initial velocity u at an angle θ with the horizontal x- axis as shown in ... So, maximum height would be, .... Projectile Motion on an inclined plane:-.

Answer 2

Answer:

Explanation:

Thank you for the reference, Maitri Gupta

Let B be the angle of projection with the inclined plane whose inclination is A and u be initial velocity.

The given condition is that particle must hit the inclined plane normally, we can use new set of coordinate axes where x-axis is along the plane and y-axis normal to the plane. The point at which the velocity along the x-axis becomes zero, the particle always hits the plane normally.

Components of acceleration due gravity along the plane = g sin A

Normal to the plane = g cos A

Similarly, component of velocity along the plane = u cos(B - A)

Normal to the plane = u sin(B - A)

Let t be the time after projection, the condition is met. We have respective equations of motion as

x = u cos (B - A)t - ½ g sin(A)t^2 …….(1)

y = u sin(B - A)t - ½ g cos(A)t^2 ……..(2)

From (1) point when x-component of velocity becomes zero is

0 = u cos (B - A) - g sin(A)t

⇒ u cos (B - A) = g sin(A)t ……..(3)

Along y-axis displacement must be zero for the particle to hit the plane. Hence, from (2) we get

0 = u sin(B - A)t - ½ g cos(A)t^2

⇒ u sin(B - A) = ½ g cos(A)t ……(4)

Dividing (4) by (3) we get

2 tan (B - A) = cot A

Angle of projection from the inclined plane = Arctan (½ cot A)

Hope this helps.