A body is dropped from a height of 16 m. The body strikes the ground and losses 25% of its velocity. The body rebounds to a height of

Options

A. 12 m
B. 9 m
C. 4 m
D. 8 m

Answer

To solve the problem step by step, we will follow the physics principles of motion under gravity and the conservation of energy.

### Step 1: Calculate the velocity of the body just before it hits the ground.

When the body is dropped from a height \( h = 16 \, \text{m} \), we can use the following kinematic equation:

\[ V^2 = U^2 + 2AS \]

Where:
- \( V \) = final velocity (just before hitting the ground)
- \( U \) = initial velocity (0 m/s, since it is dropped)
- \( A \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))
- \( S \) = distance fallen (16 m)

Substituting the values:

\[ V^2 = 0 + 2 \times 10 \times 16 \]
\[ V^2 = 320 \]
\[ V = \sqrt{320} = 8\sqrt{5} \, \text{m/s} \]

### Step 2: Determine the velocity after the body rebounds.

The body loses 25% of its velocity upon hitting the ground. Therefore, it retains 75% of its velocity:

\[ \text{Remaining velocity} = \frac{3}{4} \times V = \frac{3}{4} \times 8\sqrt{5} \]
\[ = 6\sqrt{5} \, \text{m/s} \]

### Step 3: Calculate the height to which the body rebounds.

When the body rebounds, we can again use the kinematic equation, but this time we will consider the upward motion where the final velocity \( V = 0 \) at the maximum height. The initial velocity \( U = 6\sqrt{5} \, \text{m/s} \) and the acceleration \( A = -g = -10 \, \text{m/s}^2 \).

Using the equation:

\[ V^2 = U^2 + 2AS \]

Substituting the known values:

\[ 0 = (6\sqrt{5})^2 + 2(-10)S \]
\[ 0 = 180 - 20S \]
\[ 20S = 180 \]
\[ S = \frac{180}{20} = 9 \, \text{m} \]

### Conclusion

The body rebounds to a height of **9 meters**.

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