In quick succession, a large number of balls are throun 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 mumber of the thrown up per second (g=10 m s^(-2)).

Explanation

To solve the problem of how many balls are thrown up per second when the maximum height is 5 m, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the known values:**
- Maximum height (S) = 5 m
- Acceleration due to gravity (g) = 10 m/s² (acting downwards)

2. **Use the third equation of motion to find the initial velocity (U):**
The third equation of motion states:
\[ V^2 - U^2 = 2gS \]
At the maximum height, the final velocity (V) is 0. Therefore, we can rewrite the equation as:
\[ 0 - U^2 = -2 \times g \times S \]
Substituting the known values:
\[ -U^2 = -2 \times 10 \times 5 \]
Simplifying this gives:
\[ U^2 = 100 \]
Taking the square root:
\[ U = 10 \text{ m/s} \]

3. **Calculate the time taken to reach maximum height (T):**
We will use the first equation of motion:
\[ V = U - gT \]
At maximum height, V = 0, so:
\[ 0 = 10 - 10T \]
Rearranging gives:
\[ 10T = 10 \implies T = 1 \text{ second} \]

4. **Determine the number of balls thrown per second:**
Since the next ball is thrown when the previous ball reaches its maximum height, and it takes 1 second for the ball to reach its maximum height, we can conclude:
- Number of balls thrown per second = 1 ball.

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

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