Solution:



Let's use the kinematic equation:

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is acceleration due to gravity and s is the displacement.

At the maximum height, the final velocity is 0, so we can write:

0² = u² - 2gh

where g is acceleration due to gravity and h is the maximum height.

Solving for h, we get:

h = u²/2g

The time taken to reach the maximum height can be found using another kinematic equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is acceleration due to gravity and t is time taken.

At the maximum height, the final velocity is 0, so we can write:

0 = u - gt

Solving for t, we get:

t = u/g



Using the same kinematic equation:

v² = u² + 2as

we can find the velocity of the ball at any point during its trajectory.

Let's find the velocity when the ball is at a height of h/2.

The initial velocity is u and the displacement is h/2.

The acceleration due to gravity is -g (since it acts in the opposite direction to the motion of the ball).

So we can write:

v² = u² + 2(-g)(h/2 - 0)

v² = u² - gh

At this point, the velocity is half of the maximum velocity. So we can write:

(v/2)² = u²/2 - gh/2

Solving for h/2, we get:

h/2 = u²/8g



Using the kinematic equation:

v = u + at

we can find the time taken to reach a certain velocity.

Let's find the time taken to reach a velocity of u/2.

The initial velocity is u and the final velocity is u/2.

The acceleration due to gravity is -g (since it acts in the opposite direction to the motion of the ball).

So we can write:

u/2 = u - gt

Solving for t, we get:

t = u/2g



Using the kinematic equation:

v² = u² + 2as

we can find the position of the ball at any point during its trajectory.

Let's find the position when the ball's velocity is half of the maximum velocity.

The initial velocity is u and the final velocity is u/2.

The acceleration due to gravity is -g (since it acts in the opposite direction to the motion of the ball).

Let's assume that the position when the velocity is half of the maximum velocity is x.

So we can write:

(u/2)² = u² - 2g(x - 0)

u²/4 = u² - 2gx

Solving for x, we get: