A boy sees a ball go up and then down through a window 2.45 m high. If the total time that ball is in sight in 1s, the height above the window the ball rises is approximately

Options

A. `2.45 m`
B. `4.9 m`
C. `0.3 m`
D. `0.49 m`

Answer

To solve the problem step by step, we will analyze the motion of the ball as it goes up and then down through the window.

### Step 1: Understand the problem
The ball is seen going up and down through a window that is 2.45 m high. The total time the ball is in sight is 1 second. We need to find the height above the window that the ball rises.

### Step 2: Determine the time to reach maximum height
Since the total time for the ball to go up and down is 1 second, the time taken to reach the maximum height is half of that:
\[ t_{\text{up}} = \frac{1}{2} \text{ seconds} = 0.5 \text{ seconds} \]

### Step 3: Use the kinematic equation
We can use the kinematic equation for the distance traveled under uniform acceleration:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \(s\) is the distance traveled (which is the height of the window, 2.45 m),
- \(u\) is the initial velocity,
- \(a\) is the acceleration (which is \(-g = -10 \, \text{m/s}^2\) since it is going upwards),
- \(t\) is the time (0.5 seconds).

Substituting the known values:
\[ 2.45 = u \cdot 0.5 - \frac{1}{2} \cdot 10 \cdot (0.5)^2 \]

### Step 4: Simplify the equation
Calculating the second term:
\[ \frac{1}{2} \cdot 10 \cdot (0.5)^2 = \frac{10}{2} \cdot \frac{1}{4} = \frac{10}{8} = 1.25 \]
Now substituting this back into the equation:
\[ 2.45 = 0.5u - 1.25 \]

### Step 5: Solve for initial velocity \(u\)
Rearranging the equation:
\[ 0.5u = 2.45 + 1.25 \]
\[ 0.5u = 3.70 \]
\[ u = \frac{3.70}{0.5} = 7.4 \, \text{m/s} \]

### Step 6: Calculate the maximum height above the window
Using the formula for maximum height:
\[ h = \frac{u^2}{2g} \]
Substituting \(u = 7.4 \, \text{m/s}\) and \(g = 10 \, \text{m/s}^2\):
\[ h = \frac{(7.4)^2}{2 \cdot 10} = \frac{54.76}{20} = 2.738 \, \text{m} \]

### Step 7: Find the total height above the ground
The total height above the ground is the height of the window plus the height above the window:
\[ \text{Total height} = 2.738 + 2.45 = 5.188 \, \text{m} \]

### Step 8: Conclusion
The height above the window that the ball rises is approximately:
\[ h \approx 2.738 \, \text{m} \]