Just before striking the ground a 5.0 kg body has a kinetic energy of 980 J. IF friction is ignored, from what height was it dropped ?

Options

A. 9.8 m
B. 50 m
C. 20.0 m
D. 24 m

Answer

To solve the problem, we will use the principle of conservation of energy, which states that the potential energy (PE) at the height from which the body is dropped is converted into kinetic energy (KE) just before it strikes the ground.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the body (m) = 5.0 kg
- Kinetic energy just before striking the ground (KE) = 980 J
- Acceleration due to gravity (g) = 9.8 m/s² (standard value)

2. **Write the Formula for Kinetic Energy:**
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
However, in this case, we will relate kinetic energy to potential energy.

3. **Write the Formula for Potential Energy:**
The potential energy (PE) at height (h) is given by:
\[ PE = mgh \]

4. **Apply Conservation of Energy:**
Since we are ignoring friction, the kinetic energy just before hitting the ground is equal to the potential energy at the height from which it was dropped:
\[ KE = PE \]
Therefore:
\[ 980 J = mgh \]

5. **Substitute the Known Values:**
Substitute the values of mass (m = 5 kg) and acceleration due to gravity (g = 9.8 m/s²) into the equation:
\[ 980 = 5 \cdot 9.8 \cdot h \]

6. **Solve for Height (h):**
Rearranging the equation to solve for h:
\[ h = \frac{980}{5 \cdot 9.8} \]
Calculate the denominator:
\[ 5 \cdot 9.8 = 49 \]
Now substitute back:
\[ h = \frac{980}{49} \]
Performing the division:
\[ h = 20 \text{ meters} \]

### Final Answer:
The height from which the body was dropped is **20 meters**.