A ball of mass 1kg is dropped from a height of 5m. (a) Find the kinetic of the ball just before it reaches the ground (b) What is the speed at this instant?

Explanation

To solve the problem step by step, we will break it down into two parts: (a) finding the kinetic energy of the ball just before it reaches the ground, and (b) calculating the speed of the ball at that instant.

### Step-by-Step Solution:

**(a) Finding the Kinetic Energy:**

1. **Understand the Energy Conversion:**
When the ball is dropped from a height, its potential energy at the height will convert into kinetic energy just before it hits the ground.

2. **Formula for Potential Energy (PE):**
The potential energy (PE) of an object at height \( h \) is given by the formula:
\[ PE = mgh \]
where:
- \( m \) = mass of the ball (1 kg)
- \( g \) = acceleration due to gravity (approximately 10 m/s²)
- \( h \) = height (5 m)

3. **Substituting the Values:**
\[ PE = 1 \, \text{kg} \times 10 \, \text{m/s}^2 \times 5 \, \text{m} = 50 \, \text{Joules} \]

4. **Kinetic Energy Just Before Hitting the Ground:**
Since all potential energy converts to kinetic energy just before the ball hits the ground, we have:
\[ KE = PE = 50 \, \text{Joules} \]

**(b) Finding the Speed:**

1. **Formula for Kinetic Energy (KE):**
The kinetic energy of an object is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \( m \) = mass of the ball (1 kg)
- \( v \) = speed of the ball

2. **Setting Kinetic Energy Equal to 50 Joules:**
\[ \frac{1}{2} mv^2 = 50 \]
Substituting \( m = 1 \, \text{kg} \):
\[ \frac{1}{2} \times 1 \, \text{kg} \times v^2 = 50 \]

3. **Solving for \( v^2 \):**
\[ \frac{1}{2} v^2 = 50 \implies v^2 = 100 \]

4. **Finding the Speed \( v \):**
\[ v = \sqrt{100} = 10 \, \text{m/s} \]

### Final Answers:
- (a) The kinetic energy of the ball just before it reaches the ground is **50 Joules**.
- (b) The speed of the ball at this instant is **10 m/s**.