A ball is thrown up and attains a maximum height of 100 m .Its initial speed was

Options

A. `9.8ms^(-1)`
B. `44.2ms^(-1)`
C. `19.6ms^(-1)`
D. none of these

Answer

To find the initial speed of a ball thrown upwards that reaches a maximum height of 100 meters, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the known values
- Maximum height (s) = 100 m
- Final velocity (v) at maximum height = 0 m/s (the ball stops momentarily at the highest point)
- Acceleration (a) = -g = -9.8 m/s² (the acceleration due to gravity acts downwards)

### Step 2: Use the third equation of motion
The third equation of motion is given by:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration
- \( s \) = displacement

### Step 3: Substitute the known values into the equation
Since we know that at the maximum height, the final velocity \( v = 0 \):
\[ 0 = u^2 + 2(-9.8)(100) \]

### Step 4: Simplify the equation
This simplifies to:
\[ 0 = u^2 - 1960 \]
Rearranging gives:
\[ u^2 = 1960 \]

### Step 5: Solve for the initial velocity (u)
Taking the square root of both sides:
\[ u = \sqrt{1960} \]

### Step 6: Calculate the value
Calculating the square root:
\[ u \approx 44.27 \, \text{m/s} \]

### Conclusion
The initial speed of the ball is approximately **44.27 m/s**.

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