To show that $F = ma$ is a special case of Newton's second law, we start with the definition of force and motion:

• According to Newton's second law:

$F = \frac{d(mv)}{dt}$

If mass $m$ is constant, we can factor it out, so:

$F = m \frac{dv}{dt} = ma$

Thus, $F = ma$ is derived under the condition of constant mass, which confirms it as a special case of Newton's second law.

Apparatus

• Two vectors on a board
• A protractor
• A ruler

Method

1. Draw the first vector on a horizontal plane.
2. Use the protractor to measure the angle of the second vector from the first vector.
3. Draw the second vector at the measured angle from the tip of the first vector.
4. Complete the triangle by drawing the resultant vector from the tail of the first vector to the tip of the second vector.

Observation

• Measure the length of the resultant vector and the angle it makes with the horizontal to find its magnitude and direction.

According to Newton's first law, an object in motion will continue in motion unless acted upon by an external force. As the ball approaches, moving towards her, she moves her hands away to avoid a sudden impact, allowing her to control the catch better. This action also reduces the risk of injury by accommodating the ball's momentum.

We use the kinematic equation:

$s = ut + \frac{1}{2} a t^2$

Where:

• Initial velocity $u = 28 \text{ m/s} \times \sin(45^{\text{o}}) = 28 \text{ m/s} \times \frac{\sqrt{2}}{2} \approx 19.8 \text{ m/s}$ (vertical component)
• Displacement $s = 16.0 - 1.6 = 14.4 \text{ m}$
• Acceleration $a = -9.8 \text{ m/s}^2$ (due to gravity)

Solving for time $t$ gives:

By substituting values, we find:

The total time of flight can be calculated using:

t = $\frac{u + \sqrt{u^2 + 2as}}{a}$. After calculation, it gives $4.04 \text{ s}$.

Using the horizontal component of the initial velocity:

$u_{horizontal} = u \cdot \cos(45^{\text{o}}) = 28 \text{ m/s} \cdot \frac{\sqrt{2}}{2} \approx 19.8 \text{ m/s}$

The formula for horizontal distance is:

$s = ut\implies s = 19.8 \text{ m/s} \times 4.04 \text{ s} \approx 80 \text{ m}.$

Using the kinematic equation:

$s = \frac{1}{2} a t^2 + u t$

For the vertical motion: $s = u_{vertical}t + \frac{1}{2}(-9.8)(t^2)$

Where $u = 19.8 \text{ m/s}$ and substituting $t = 4.04 \text{ s}$, we can calculate:

$s = 19.8 \cdot 4.04 + \frac{1}{2}(-9.8) \cdot (4.04)^2$

This results in the maximum height being: $s = 20 + 1.6= 21.6 ext{m}$

The diagram should depict:

• Velocity vector $v$ pointing horizontally (magnitude being the horizontal component).
• Acceleration vector $a$ pointing downwards (due to gravity).
• The force vector $F$ acting on the ball at that position, which corresponds to its weight.

Label these appropriately with vectors showing their directions.