A particle is released on the smooth inside wall of cylinder tank at A with velocity v0 which makes an angle β with horizontal tangent. The radius of the tank is r. Find the maximum height above point A reached by the particle, the magnitude of the velocity, and the angle ϴ formed by the velocity vector and the horizontal tangent at this point.

To find the maximum height reached by the particle, we need to consider the forces acting on it. The only force acting is the gravitational force.

Let's denote the maximum height as h.

At the highest point, the velocity of the particle is perpendicular to the radius of the cylinder. This means that the centripetal force required at that point is provided completely by the gravitational force acting on the particle.

Considering these forces, we can set up the following equation:

m * g = m * v^2 / r

where m is the mass of the particle, g is the acceleration due to gravity, v is the magnitude of the velocity, and r is the radius of the cylinder.

We can cancel out the mass (m) on both sides of the equation:

g = v^2 / r

Now we can solve for v:

v^2 = g * r

v = sqrt(g * r)

To find the maximum height (h), we can use the kinetic energy equation:

1/2 * m * v^2 = m * g * h

Canceling out the mass (m), we have:

1/2 * v^2 = g * h

Substituting the value of v^2 from the previous equation:

1/2 * (g * r) = g * h

Simplifying the equation:

h = r / 2

Therefore, the maximum height above point A reached by the particle is equal to half the radius of the tank.

To find the angle ϴ formed by the velocity vector and the horizontal tangent at this point, we can use basic trigonometry. Since the particle is released with an angle β with the horizontal tangent, the angle ϴ formed will be equal to β.

Therefore, the angle ϴ formed by the velocity vector and the horizontal tangent at the highest point is equal to β.