Answer: t any time instant t, the speed of ball A relative to ball B is u.

Explanation:To answer the questions, let's consider the motion of ball A and ball B separately.

For ball A:

The initial velocity of ball A is u (thrown up vertically).

The acceleration due to gravity acting on ball A is -g (negative because it acts in the opposite direction of motion).

The displacement of ball A at any time t can be calculated using the equation:

s_A = ut - (1/2)gt^2 (where s_A is the displacement of ball A)

For ball B:

Ball B is released from rest at a height h.

The acceleration due to gravity acting on ball B is -g.

The displacement of ball B at any time t can be calculated using the equation:

s_B = h - (1/2)gt^2 (where s_B is the displacement of ball B)

Now, let's answer the questions:

At the moment of collision, ball B will be at the same height as ball A, since both balls were released at the same time and ball B was released from a height h. Therefore, the position of ball B at the moment of collision is h.

The relative acceleration of A with respect to B can be calculated by subtracting the acceleration of ball B from the acceleration of ball A. Since both balls experience the same acceleration due to gravity (-g), the relative acceleration of A with respect to B is 0.

The speed of A relative to B can be calculated by subtracting the velocity of ball B from the velocity of ball A. The velocity of ball A at any time t can be calculated as:

v_A = u - gt (where v_A is the velocity of ball A)

The velocity of ball B at any time t is:

v_B = -gt (where v_B is the velocity of ball B)

Therefore, the speed of A relative to B is:

speed_A/B = |v_A - v_B| = |u - gt - (-gt)| = |u| = u

So, at any time instant t, the speed of ball A relative to ball B is u.

Verified answer

Answer:

Explanation:

9. Let's first find the time of collision. The position of ball A at any time t is given by the equation yA = ut - 0.5gt^2, where yA is the height of ball A above its starting point. This equation comes from the kinematic equation for an object undergoing constant acceleration, where the initial velocity is u and the acceleration is -g (due to gravity).

The position of ball B at any time t is given by the equation yB = h - 0.5gt^2, where yB is the height of ball B above the starting point of ball A. This equation also comes from the kinematic equation for an object undergoing constant acceleration, where the initial velocity is 0 (since ball B is released from rest) and the acceleration is -g (due to gravity).

At the moment of collision, yA = yB. Setting these two equations equal to each other and solving for t, we get:

ut - 0.5gt^2 = h - 0.5gt^2

ut = h

t = h/u

Substituting this value of t into either equation for yA or yB, we find that at the moment of collision, the position of B is at a distance h - 0.5g(h/u)^2 from its starting point.

10. The acceleration of both balls is due to gravity and is equal in magnitude and opposite in direction. Therefore, the magnitude of relative acceleration of A with respect to B is 0.

11. The speed of ball A at any time t is given by the derivative of its position equation: vA = u - gt. This comes from taking the derivative of yA = ut - 0.5gt^2 with respect to t.

The speed of ball B at any time t is given by the derivative of its position equation: vB = -gt. This comes from taking the derivative of yB = h - 0.5gt^2 with respect to t.

Therefore, at any time instant t, the speed of A relative to B is vA - vB = u.