A man of 60kg gains 1000 cal of heat by eating 5 mangoes. His efficiency is 56%. To what height he can jump by using this energy?

Options

A. 4m
B. 20m
C. 28m
D. 0.2m

Explanation

To solve the problem step by step, we need to determine how high the man can jump using the energy he gains from eating the mangoes, considering his efficiency.

### Step 1: Convert Calories to Joules
The first step is to convert the heat energy gained from calories to joules. We know that:
1 calorie = 4.184 joules.

Given that the man gains 1000 calories:
\[ \text{Energy in joules} = 1000 \, \text{cal} \times 4.184 \, \text{J/cal} = 4184 \, \text{J} \]

### Step 2: Calculate the Useful Energy
Since the man's efficiency is 56%, we need to calculate the useful energy that he can actually use for jumping:
\[ \text{Useful energy} = \text{Total energy} \times \text{Efficiency} \]
\[ \text{Useful energy} = 4184 \, \text{J} \times 0.56 = 2343.04 \, \text{J} \]

### Step 3: Use the Potential Energy Formula
The potential energy (PE) gained when jumping to a height \( h \) is given by the formula:
\[ PE = mgh \]
Where:
- \( m \) = mass of the man = 60 kg
- \( g \) = acceleration due to gravity = 9.81 m/s²
- \( h \) = height in meters

We can rearrange the formula to solve for height \( h \):
\[ h = \frac{PE}{mg} \]

### Step 4: Substitute the Values
Now, substituting the values we have:
\[ h = \frac{2343.04 \, \text{J}}{60 \, \text{kg} \times 9.81 \, \text{m/s}^2} \]
\[ h = \frac{2343.04}{588.6} \approx 3.98 \, \text{m} \]

### Step 5: Round the Height
Rounding 3.98 m gives approximately 4 m. Therefore, the height to which the man can jump is:
\[ \text{Height} \approx 4 \, \text{m} \]

### Final Answer
The man can jump to a height of approximately **4 meters**.

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