A 0.5 kg ball is thrown up with an initial speed 14 m/s and reaches a maximum height of 8.0 m . How much energy is dissipated by air drag acting on the ball during the ascent

Options

A. 19.6 Joule
B. 4.9 Joule
C. 10 Joule
D. 9.8 Joule

Answer

To solve the problem of how much energy is dissipated by air drag acting on a ball thrown upwards, we can follow these steps:

### Step 1: Calculate the initial kinetic energy (KE_initial)
The initial kinetic energy of the ball can be calculated using the formula:
\[ KE_{\text{initial}} = \frac{1}{2} mv^2 \]
Where:
- \( m = 0.5 \, \text{kg} \) (mass of the ball)
- \( v = 14 \, \text{m/s} \) (initial speed)

Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 0.5 \, \text{kg} \times (14 \, \text{m/s})^2 = \frac{1}{2} \times 0.5 \times 196 = 49 \, \text{J} \]

### Step 2: Calculate the potential energy (PE_final) at maximum height
The potential energy at the maximum height can be calculated using the formula:
\[ PE_{\text{final}} = mgh \]
Where:
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 8 \, \text{m} \) (maximum height)

Substituting the values:
\[ PE_{\text{final}} = 0.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 8 \, \text{m} = 0.5 \times 9.8 \times 8 = 39.2 \, \text{J} \]

### Step 3: Calculate the energy dissipated by air drag
The energy dissipated by air drag can be found by calculating the difference between the initial kinetic energy and the potential energy at maximum height:
\[ \text{Energy dissipated} = KE_{\text{initial}} - PE_{\text{final}} \]
Substituting the values:
\[ \text{Energy dissipated} = 49 \, \text{J} - 39.2 \, \text{J} = 9.8 \, \text{J} \]

### Conclusion
The energy dissipated by air drag acting on the ball during the ascent is:
\[ \text{Energy dissipated} = 9.8 \, \text{J} \]

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