An elevator without a ceiling is ascending up with an acceleration of `5 ms^-2.` A boy on the elevator shoots a ball in vertical upward direction from a height of 2 m above the floor of elevator. At this instant the elevator is moving up with a velocity of `10 ms^-1` and floor of the elevator is at a height of 50 m from the ground. The initial speed of the ball is `15 ms^-1` with respect to the elevator. Consider the duration for which the ball strikes the floor of elevator in answering following questions. (`g=10 ms^-2`)
2. The maximum height reached by ball, as measured from the ground would be

Options

A. 73.65 m
B. 116.25 m
C. 82.56m
D. 63.25 m

Explanation

To find the maximum height reached by the ball as measured from the ground, we will follow these steps:

### Step 1: Determine the initial velocity of the ball with respect to the ground
The ball is shot from the elevator, which is moving upwards. The initial speed of the ball with respect to the elevator is given as \(15 \, \text{m/s}\). The elevator itself is moving upwards with a velocity of \(10 \, \text{m/s}\). Therefore, the initial velocity of the ball with respect to the ground can be calculated as follows:

\[ u = v_{\text{ball}} + v_{\text{elevator}} = 15 \, \text{m/s} + 10 \, \text{m/s} = 25 \, \text{m/s} \]

### Step 2: Calculate the time taken to reach maximum height
The time taken to reach maximum height can be calculated using the formula:

\[ t_{\text{max}} = \frac{u}{g} \]

Where \(g = 10 \, \text{m/s}^2\) (acceleration due to gravity). Substituting the values:

\[ t_{\text{max}} = \frac{25 \, \text{m/s}}{10 \, \text{m/s}^2} = 2.5 \, \text{s} \]

### Step 3: Determine the time until the ball hits the floor of the elevator
From the previous calculations, we know that the ball will collide with the floor of the elevator after \(2.13 \, \text{s}\). Since \(2.13 \, \text{s} < 2.5 \, \text{s}\), the ball will not reach its maximum height before it hits the floor of the elevator.

### Step 4: Calculate the maximum height reached by the ball before hitting the floor
To find the height attained by the ball when it hits the floor of the elevator, we can use the following kinematic equation:

\[ H = H_0 + ut - \frac{1}{2}gt^2 \]

Where:
- \(H_0\) is the initial height of the ball above the ground (50 m from the ground + 2 m above the floor of the elevator = 52 m).
- \(u = 25 \, \text{m/s}\) (initial velocity of the ball).
- \(t = 2.13 \, \text{s}\) (time until it hits the floor).
- \(g = 10 \, \text{m/s}^2\).

Substituting the values:

\[ H = 52 \, \text{m} + (25 \, \text{m/s} \times 2.13 \, \text{s}) - \frac{1}{2} \times 10 \, \text{m/s}^2 \times (2.13 \, \text{s})^2 \]

Calculating each term:

1. \(25 \, \text{m/s} \times 2.13 \, \text{s} = 53.25 \, \text{m}\)
2. \(\frac{1}{2} \times 10 \, \text{m/s}^2 \times (2.13 \, \text{s})^2 = 5 \times 4.5369 = 22.6845 \, \text{m}\)

Now substituting these back into the height equation:

\[ H = 52 \, \text{m} + 53.25 \, \text{m} - 22.6845 \, \text{m} \]

Calculating:

\[ H = 52 + 53.25 - 22.6845 = 82.5655 \, \text{m} \]

### Final Answer
The maximum height reached by the ball, as measured from the ground, is approximately \(82.57 \, \text{m}\).

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