A very large number of balls are thrown vertically upwards in quick successions in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is 5m, then number of balls thrown per minute is (take `g=10m//s^(2)`)

Options

A. 82
B. 120
C. 40
D. 60

Explanation

To solve the problem, we need to determine how many balls can be thrown in one minute, given that each ball reaches a maximum height of 5 meters and is thrown when the previous one is at its maximum height.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Maximum height (h) = 5 m
- Acceleration due to gravity (g) = 10 m/s²

2. **Use the Equation of Motion:**
We can use the kinematic equation that relates initial velocity (u), final velocity (v), acceleration (a), and displacement (s):
\[ v^2 = u^2 + 2as \]
Here, at the maximum height, the final velocity (v) is 0, and the displacement (s) is the maximum height (h). The acceleration (a) is -g (since gravity acts downwards).

Plugging in the values:
\[ 0 = u^2 - 2gh \]
Rearranging gives:
\[ u^2 = 2gh \]

3. **Calculate Initial Velocity (u):**
Substituting the values of g and h:
\[ u^2 = 2 \times 10 \, \text{m/s}^2 \times 5 \, \text{m} = 100 \, \text{m}^2/\text{s}^2 \]
Taking the square root:
\[ u = \sqrt{100} = 10 \, \text{m/s} \]

4. **Determine the Time to Reach Maximum Height:**
We can use another kinematic equation to find the time (t) taken to reach the maximum height:
\[ v = u + at \]
Here, v = 0 (at maximum height), u = 10 m/s, and a = -g = -10 m/s².
Rearranging gives:
\[ 0 = 10 - 10t \]
Solving for t:
\[ 10t = 10 \implies t = 1 \, \text{s} \]

5. **Calculate the Number of Balls Thrown in One Minute:**
Since one ball is thrown every second, in one minute (60 seconds), the number of balls thrown will be:
\[ \text{Number of balls} = 60 \, \text{seconds} \times 1 \, \text{ball/second} = 60 \, \text{balls} \]

### Final Answer:
The number of balls thrown per minute is **60**.

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