A 50.7-g golf ball is driven from the tee with an initial speed of 51.2 m/s and rises to a height of 21.8 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 6.55 m below its highest point?

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To determine the kinetic energy of the golf ball at its highest point and its speed when it is 6.55 m below its highest point, we can use the principles of conservation of energy.

(a) To find the kinetic energy of the ball at its highest point, we need to consider the initial kinetic energy and the potential energy at that point. Since the ball rises to a height of 21.8 m, the potential energy at its highest point is given by:

Potential energy = mgh

where m is the mass of the ball (50.7 g = 0.0507 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (21.8 m).

Potential energy = (0.0507 kg)(9.8 m/s^2)(21.8 m) = 10.972 J

Since there is no air resistance, the total mechanical energy of the system is conserved. Therefore, at its highest point, the kinetic energy of the ball is equal to the initial kinetic energy. The initial kinetic energy is given by:

Kinetic energy = (1/2)mv^2

where m is the mass of the ball (0.0507 kg) and v is the initial speed (51.2 m/s).

Kinetic energy = (1/2)(0.0507 kg)(51.2 m/s)^2 = 66.170 J

Therefore, the kinetic energy of the ball at its highest point is 66.170 J.

(b) To find the speed when the ball is 6.55 m below its highest point, we need to consider the conservation of mechanical energy again. The total mechanical energy at the highest point is equal to the total mechanical energy at 6.55 m below.

Let's call the speed at this point v'.

The total mechanical energy at 6.55 m below its highest point is given by:

Mechanical energy = Kinetic energy + Potential energy

The potential energy is given by:

Potential energy = mgh

where m is the mass of the ball (0.0507 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (6.55 m).

Potential energy = (0.0507 kg)(9.8 m/s^2)(6.55 m) = 3.181 J

Since the total mechanical energy is conserved, we can set up the equation:

(1/2)mv^2 + mgh = (1/2)mv'^2 + mgh

Simplifying the equation, we get:

(1/2)mv^2 = (1/2)mv'^2 + mgh

Substituting the known values:

(1/2)(0.0507 kg)(51.2 m/s)^2 = (1/2)(0.0507 kg)(v')^2 + (0.0507 kg)(9.8 m/s^2)(6.55 m)

Solving this equation will give us the value of v', which is the speed of the ball when it is 6.55 m below its highest point.

By solving the equation, we find:

(1/2)(0.0507 kg)(v')^2 = 16.9633 J

(v')^2 = (2(16.9633 J))/(0.0507 kg)

v' = √((2(16.9633 J))/(0.0507 kg))

Using a calculator, we find:

v' ≈ 34.2 m/s

Therefore, the speed of the ball when it is 6.55 m below its highest point is approximately 34.2 m/s.