A cricketer hits a cricket ball of 160 g straight up into the air. The cricket ball has an init velocity of $$ 20 \mathrm{~m} \mathrm{~s}^{-1} $$ vertically upwards. 1.1 Draw a free body diagram for the cricket ball. 1.2 What height does the ball reach before it stops to fall back to the ground? 1.3 How long has the ball been in the air for? 1.4 Calculate the linear momentum of the ball at the moment it reaches the ground.

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Let's tackle each part of the problem step-by-step.

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### **1.1 Free-Body Diagram for the Cricket Ball**

A **free-body diagram** illustrates all the forces acting on an object. For the cricket ball in this scenario:

- **Weight (Gravitational Force)**: This force acts **downward** and is given by $$ W = mg $$, where:
  - $$ m = 0.160 $$ kg (mass of the ball)
  - $$ g = 9.81 \, \text{m/s}² $$ (acceleration due to gravity)

Assuming air resistance is negligible, the **only** force acting on the cricket ball during its flight is its weight.

**Diagram Description:**
- Draw the cricket ball as a dot or small circle.
- Draw an arrow pointing **downward** from the center of the ball labeled $$ W = mg $$.

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### **1.2 Maximum Height Reached by the Ball**

To find the maximum height ($$ h $$) the ball reaches before stopping, we'll use the **kinematic equation**:

$$
v^2 = u^2 + 2as
$$

Where:
- $$ v = 0 \, \text{m/s} $$ (final velocity at the highest point)
- $$ u = 20 \, \text{m/s} $$ (initial upward velocity)
- $$ a = -g = -9.81 \, \text{m/s}² $$ (acceleration due to gravity, negative since it's acting downward)
- $$ s = h $$ (displacement, which is the maximum height)

Rearranging the equation to solve for $$ h $$:

$$
0 = (20)^2 + 2(-9.81)h \
400 = 19.62h \
h = \frac{400}{19.62} \
h \approx 20.39 \, \text{meters}
$$

**Answer:** The ball reaches a maximum height of approximately **20.39 meters** before it starts to fall back to the ground.

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### **1.3 Total Time the Ball is in the Air**

The total time ($$ T $$) the ball spends in the air consists of:
- **Time to ascend** ($$ t_{\text{up}} $$)
- **Time to descend** ($$ t_{\text{down}} $$)

Since the motion upwards and downwards are symmetrical (neglecting air resistance):

$$
T = 2 \times t_{\text{up}}
$$

First, find $$ t_{\text{up}} $$ using the formula:

$$
v = u + at
$$

At the highest point, $$ v = 0 $$:

$$
0 = 20 - 9.81 \times t_{\text{up}} \
t_{\text{up}} = \frac{20}{9.81} \
t_{\text{up}} \approx 2.04 \, \text{seconds}
$$

Thus,

$$
T = 2 \times 2.04 \approx 4.08 \, \text{seconds}
$$

**Answer:** The ball is in the air for approximately **4.08 seconds**.

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### **1.4 Linear Momentum of the Ball Upon Returning to the Ground**

**Linear momentum ($$ p $$)** is given by:

$$
p = m \times v
$$

When the ball returns to the ground:
- Its speed is equal in magnitude to the initial speed (20 m/s) but in the **opposite direction** (downward).
- Therefore, $$ v = -20 \, \text{m/s} $$ (negative sign indicates direction)

Calculating the momentum:

$$
p = 0.160 \, \text{kg} \times (-20 \, \text{m/s}) \
p = -3.2 \, \text{kg·m/s}
$$

**Answer:** Upon reaching the ground, the ball has a linear momentum of **–3.2  kg·m/s** (directed downward).

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**1.1 Free-Body Diagram:** Draw the cricket ball with a downward arrow labeled $$ W = 0.160 \times 9.81 $$ (weight). **1.2 Maximum Height:** The ball reaches approximately **20.39 meters** before stopping. **1.3 Total Time in Air:** The ball is in the air for about **4.08 seconds**. **1.4 Linear Momentum Upon Returning:** The ball has a momentum of **–3.2  kg·m/s** downward when it hits the ground.

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