a ball of mass 200 g is thrown vertically upwards with a velocity of 20ms-1 ignoring air resistance.

a) calculate the kinetic energy of the ball as it is released
b)calculate the maximum height reached by the ball
c) state the velocity of the ball when it returns to its starting position
d)state which conservation principle you used to obtain your answers to questions (b) and (C)
e) if the ball hit the ground 2 metres below the point from which it was thrown, calculate the impulse and impulse force of the 1 millisecond impact.

a) We can calculate the kinetic energy of the ball as it is released using the formula:

Kinetic energy (KE) = 1/2 * mass * velocity^2

Given:
mass (m) = 200 g = 0.2 kg
velocity (v) = 20 m/s

Using the formula, substituting the given values:

KE = 1/2 * 0.2 kg * (20 m/s)^2
= 1/2 * 0.2 kg * 400 m^2/s^2
= 40 J

Therefore, the kinetic energy of the ball as it is released is 40 Joules (J).

b) To calculate the maximum height reached by the ball, we can use the conservation of energy principle. At maximum height, all of the initial kinetic energy will be converted into potential energy.

Potential energy (PE) = mass * gravity * height

Given:
mass (m) = 200 g = 0.2 kg
gravity (g) = 9.8 m/s^2 (approx.)

Using the formula, substituting the given values:

PE = 0.2 kg * 9.8 m/s^2 * height

At maximum height, the velocity (v) will be 0 m/s, so the kinetic energy (KE) will be 0 J.

0 J = 1/2 * 0.2 kg * 0 m/s^2
= 0 J

Thus, at maximum height, all of the initial kinetic energy is converted into potential energy:

PE = KE
0.2 kg * 9.8 m/s^2 * height = 40 J

Solving for height:

height = 40 J / (0.2 kg * 9.8 m/s^2)
= 40 J / 1.96 N
= 20.41 m

Therefore, the maximum height reached by the ball is 20.41 meters (m).

c) The velocity of the ball when it returns to its starting position can be determined using the principle of conservation of mechanical energy. The total mechanical energy of a system remains constant if no external forces (like air resistance) are acting on it.

Since the ball returns to its starting position, its potential energy will be equal to the initial potential energy (at the start) and its velocity will be equal to the initial velocity (20 m/s) but in the opposite direction.

Therefore, the velocity of the ball when it returns to its starting position is -20 m/s.

d) We used the conservation of energy principle to obtain our answers to questions (b) and (c). This principle states that the total mechanical energy (sum of kinetic and potential energy) of a system remains constant, without any external forces acting on it.

e) To calculate the impulse and impulse force during the impact with the ground, we need the change in momentum of the ball.

Change in momentum (Δp) = Final momentum - Initial momentum

The initial momentum of the ball is given by its mass (m) multiplied by the initial velocity (v):
Initial momentum = m * v

Given:
mass (m) = 0.2 kg
initial velocity (v) = 20 m/s

Initial momentum = 0.2 kg * 20 m/s
= 4 kg*m/s

Since the ball hits the ground 2 meters below the point from which it was thrown, it gains a downward velocity.

Final velocity (vf) = -20 m/s (opposite direction)

The final momentum can be calculated as:
Final momentum = m * vf
= 0.2 kg * (-20 m/s)
= -4 kg*m/s

The change in momentum (Δp) is given by:
Δp = Final momentum - Initial momentum
= (-4 kg*m/s) - (4 kg*m/s)
= -8 kg*m/s

The impulse (J) and impulse force (F) can be calculated using the formula:

J = Δp
F = J / Δt

Given:
Δt = 1 millisecond = 0.001 s

Using the formula, substituting the values:

J = -8 kg*m/s
F = J / Δt
= (-8 kg*m/s) / (0.001 s)
= -8000 N

Therefore, the impulse during the 1 millisecond impact is -8 kg*m/s, and the impulse force is -8000 Newtons (N) (opposite direction).

a) To calculate the kinetic energy of the ball as it is released, you can use the formula for kinetic energy:

Kinetic energy = 0.5 * mass * velocity^2

Given that the mass of the ball is 200 g (0.2 kg) and the velocity is 20 m/s, you can substitute these values into the formula:

Kinetic energy = 0.5 * 0.2 kg * (20 m/s)^2
Kinetic energy = 0.5 * 0.2 kg * 400 m^2/s^2
Kinetic energy = 40 J

Therefore, the kinetic energy of the ball as it is released is 40 Joules.

b) To calculate the maximum height reached by the ball, you can use the concept of conservation of energy. As the ball is thrown vertically upwards, it experiences a decrease in kinetic energy and an increase in potential energy. At the maximum height, all the kinetic energy is converted into potential energy.

Initial kinetic energy = Final potential energy

Therefore,

0.5 * mass * initial velocity^2 = mass * g * height

Substituting the known values:

0.5 * 0.2 kg * (20 m/s)^2 = 0.2 kg * 9.8 m/s^2 * height
200 J = 1.96 * height
height = 200 J / 1.96
height ≈ 102.04 m

Therefore, the maximum height reached by the ball is approximately 102.04 meters.

c) When the ball returns to its starting position, its velocity will have the same magnitude as the initial velocity, but with opposite direction. This means the velocity will be -20 m/s.

d) The conservation principle used to calculate the maximum height in question (b) and the velocity in question (c) is the principle of conservation of mechanical energy. In the absence of air resistance, the total mechanical energy (kinetic energy + potential energy) remains constant throughout the motion of the ball.

e) To calculate the impulse and impulse force during the 1 millisecond impact, we need more information. Specifically, we need to know the change in velocity of the ball during impact.

a. KE = 0.5m*V^2 = 0.1*(20)^2 = 40 Joules.

b. hmax = (V^2-Vo^2) / 2g.
hmax = (0-(20)^2) / -19.6 = 20.41 m.

c. V = Vo = 20 m/s.