A ball is thrown straight up so that it reaches a height of 25m. How fast was it going when it was 5m high?



To solve this problem, we can use the principles of energy conservation or kinematic equations. Let's approach it using energy conservation.

### Step 1: Consider energy at the highest point (25 m)
At the highest point, the kinetic energy is zero (since the velocity is zero), and the potential energy is maximum.

The potential energy at the maximum height is:
\[
PE_{\text{max}} = mgh_{\text{max}} = mg \times 25
\]
where:
- \(m\) is the mass of the ball,
- \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)),
- \(h_{\text{max}} = 25 \, \text{m}\).

### Step 2: Consider energy when the ball is at 5 m
When the ball is at 5 m, it has both kinetic energy and potential energy.

The potential energy at 5 m height is:
\[
PE_5 = mgh_5 = mg \times 5
\]

The kinetic energy at 5 m height is:
\[
KE_5 = \frac{1}{2} mv^2
\]
where \(v\) is the speed of the ball at 5 m height.

### Step 3: Apply the principle of conservation of mechanical energy
The total mechanical energy (sum of potential and kinetic energy) is conserved. Therefore, the total energy at the highest point is equal to the total energy at 5 m height:
\[
PE_{\text{max}} = PE_5 + KE_5
\]

Substituting the expressions:
\[
mg \times 25 = mg \times 5 + \frac{1}{2} mv^2
\]

We can cancel out the mass \(m\) from both sides (since it appears in every term):
\[
g \times 25 = g \times 5 + \frac{1}{2} v^2
\]

Now, solve for \(v\):
\[
g \times 20 = \frac{1}{2} v^2
\]
\[
v^2 = 2 \times g \times 20
\]
\[
v^2 = 40g
\]
\[
v = \sqrt{40g}
\]
Substitute \(g = 9.8 \, \text{m/s}^2\):
\[
v = \sqrt{40 \times 9.8}
\]
\[
v = \sqrt{392}
\]
\[
v \approx 19.8 \, \text{m/s}
\]

### Final Answer
The speed of the ball when it was at a height of 5 m is approximately **19.8 m/s**.

Do you want more details or have any questions? Here are five related questions you might find interesting:

1. What is the initial speed of the ball when it was thrown upwards?
2. How long does it take for the ball to reach the maximum height?
3. How long does it take for the ball to fall back down from the maximum height?
4. What is the speed of the ball when it is 15 meters high?
5. What would be the speed at 5 meters if the maximum height was doubled?

**Tip:** Remember, energy conservation is a powerful tool in solving problems involving motion and gravity.

Learn how to calculate the speed of a ball when it reaches a height of 5m using energy conservation principles. This problem involves calculating potential and kinetic energy, suitable for advanced high school students.

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