In quick succession, a large number of balls are thrown up vertically in such a way that the next ball is thrown up when the previous ball is at the maximum height. If the maximum height is `5 m`, then find the number of the balls thrown up per second (g=10 m s^(-2)).

Explanation

To solve the problem step by step, we will use the principles of kinematics and the equations of motion.

### Step 1: Identify the given data
- Maximum height (h) = 5 m
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Use the equation of motion to find the initial velocity (u)
We know that at the maximum height, the final velocity (v) is 0. We can use the third equation of motion:

\[ v^2 = u^2 + 2as \]

Here,
- v = 0 (at maximum height)
- a = -g (since gravity acts downward)
- s = h (maximum height)

Substituting the values:

\[ 0 = u^2 - 2gh \]

Rearranging gives:

\[ u^2 = 2gh \]

Substituting g = 10 m/s² and h = 5 m:

\[ u^2 = 2 \times 10 \times 5 \]
\[ u^2 = 100 \]
\[ u = \sqrt{100} = 10 \text{ m/s} \]

### Step 3: Calculate the time taken to reach maximum height (t)
We can use the first equation of motion:

\[ v = u + at \]

At maximum height, v = 0, so:

\[ 0 = u - gt \]

Substituting u = 10 m/s and g = 10 m/s²:

\[ 0 = 10 - 10t \]

Rearranging gives:

\[ 10t = 10 \]
\[ t = 1 \text{ second} \]

### Step 4: Determine the number of balls thrown per second
Since a new ball is thrown when the previous ball reaches its maximum height, and it takes 1 second for a ball to reach the maximum height, we can conclude:

- Number of balls thrown up per second = 1

### Final Answer
The number of balls thrown up per second is **1**.

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