To solve the problem of a ball being thrown over a sphere with minimum speed, we need to consider the physics of projectile motion and the geometry of the sphere. Let's break this down step by step.

Minimum Speed of Projection

To determine the minimum speed required for the ball to just clear the sphere, we can use the concept of energy conservation and projectile motion. The ball must reach a height equal to the diameter of the sphere (2R) at its peak to clear it. The minimum speed \( v \) at which the ball must be projected can be derived from the following energy considerations:

• The potential energy at the peak height must equal the initial kinetic energy.
• At the peak height, the potential energy \( PE \) is given by \( PE = mgh \), where \( h = 2R \).
• The initial kinetic energy \( KE \) is given by \( KE = \frac{1}{2} mv^2 \).

Setting these equal gives us:

\( mgh = \frac{1}{2} mv^2 \)

By canceling \( m \) from both sides and substituting \( h \) with \( 2R \), we have:

\( g(2R) = \frac{1}{2} v^2 \)

Rearranging this equation leads to:

\( v^2 = 4gR \)

Thus, the minimum speed \( v \) required is:

\( v = \sqrt{4gR} = 2\sqrt{gR} \)

Angle of Projection

Next, we need to find the angle of projection \( \theta \) that corresponds to this minimum speed. For a projectile to reach a maximum height of \( 2R \), we can use the vertical component of the initial velocity:

\( v_y = v \sin \theta

At the peak, the vertical component of the velocity becomes zero, and we can use the following kinematic equation:

\( v_y^2 = u_y^2 - 2g h

Substituting \( h = 2R \) gives us:

\( 0 = (v \sin \theta)^2 - 2g(2R)

Rearranging this leads to:

\( (v \sin \theta)^2 = 4gR

Substituting \( v = 2\sqrt{gR} \) into the equation:

\( (2\sqrt{gR} \sin \theta)^2 = 4gR

From this, we can simplify to find \( \sin^2 \theta = 1 \), which implies:

\( \sin \theta = 1 \)

Thus, \( \theta = 90^\circ \). The cosine of this angle is:

\( \cos \theta = 0

Distance from the Sphere's Lowest Point

Finally, we need to determine the horizontal distance from the point of projection to the lowest point of the sphere. When the ball is projected at an angle of \( 90^\circ \), it travels vertically upwards. However, to find the distance from the lowest point of the sphere, we can consider the horizontal component of the motion.

Since the angle of projection is \( 90^\circ \), the horizontal distance \( d \) from the point of projection to the center of the sphere is simply the radius \( R \). Therefore, the distance from the lowest point of the sphere (which is at the bottom) to the point of projection is:

\( d = R

In summary, the answers to your questions are:

• Minimum speed of projection: \( 2\sqrt{gR} \)
• Cosine of the angle of projection: \( 0 \)
• Distance from the lowest point of the sphere: \( R \)