A ballon moves up with a velocity `5 m//s`. A stone is thrown from it with a horizontal velocity `2 m//s` relative to it. The stone hits the ground at a point `10 m` horizontally away from it. (Take `g = 10 m//s^(2)`)

Options

A. the time of flight of the stone is `5` seconds
B. height of the ballon at the time of throwing the stone is `100 m`.
C. the speed with it will hit the ground is `40 m//s`
D. the time of flight of the stone will increase if the horizontal velocity of the stone is incresed.

Explanation

To solve the problem step by step, we will analyze the motion of the balloon and the stone separately, and then combine the results to find the required quantities.

### Step 1: Understand the Motion of the Balloon
The balloon is moving upwards with a velocity of \(5 \, \text{m/s}\). This means that at the moment the stone is thrown, the balloon has a vertical velocity of \(5 \, \text{m/s}\).

### Step 2: Analyze the Horizontal Motion of the Stone
The stone is thrown horizontally with a velocity of \(2 \, \text{m/s}\) relative to the balloon. Since the balloon is moving upwards, the horizontal velocity of the stone with respect to the ground will be the same, \(2 \, \text{m/s}\).

### Step 3: Determine the Time of Flight
The stone hits the ground at a point \(10 \, \text{m}\) horizontally away from the point where it was thrown. We can use the horizontal motion to find the time of flight \(t\):

\[ \text{Horizontal distance} = \text{Horizontal velocity} \times \text{Time} \]

Given that the horizontal distance is \(10 \, \text{m}\) and the horizontal velocity is \(2 \, \text{m/s}\):

\[ 10 = 2 \times t \]

Solving for \(t\):

\[ t = \frac{10}{2} = 5 \, \text{s} \]

### Step 4: Calculate the Height from Which the Stone is Thrown
Now we need to find the height \(h\) from which the stone is thrown. We can use the vertical motion equation:

\[ h = u t + \frac{1}{2} g t^2 \]

Where:
- \(u = 5 \, \text{m/s}\) (initial vertical velocity of the stone)
- \(g = 10 \, \text{m/s}^2\) (acceleration due to gravity)
- \(t = 5 \, \text{s}\)

Substituting the values:

\[ h = 5 \times 5 + \frac{1}{2} \times 10 \times (5)^2 \]

Calculating:

\[ h = 25 + \frac{1}{2} \times 10 \times 25 \]
\[ h = 25 + 125 = 150 \, \text{m} \]

### Step 5: Calculate the Final Velocity of the Stone Just Before Hitting the Ground
The final vertical velocity \(v\) of the stone just before it hits the ground can be calculated using:

\[ v = u + g t \]

Where:
- \(u = 5 \, \text{m/s}\) (initial vertical velocity)
- \(g = 10 \, \text{m/s}^2\)
- \(t = 5 \, \text{s}\)

Substituting the values:

\[ v = 5 + 10 \times 5 \]
\[ v = 5 + 50 = 55 \, \text{m/s} \]

### Step 6: Calculate the Resultant Velocity of the Stone
The resultant velocity of the stone just before it hits the ground can be found using the Pythagorean theorem, since the stone has both horizontal and vertical components of velocity:

\[ v_{\text{resultant}} = \sqrt{(v_x)^2 + (v_y)^2} \]

Where:
- \(v_x = 2 \, \text{m/s}\) (horizontal velocity)
- \(v_y = 55 \, \text{m/s}\) (vertical velocity)

Calculating:

\[ v_{\text{resultant}} = \sqrt{(2)^2 + (55)^2} \]
\[ v_{\text{resultant}} = \sqrt{4 + 3025} = \sqrt{3029} \approx 55 \, \text{m/s} \]

### Final Answer
The height from which the stone was thrown is \(150 \, \text{m}\) and the resultant velocity just before hitting the ground is approximately \(55 \, \text{m/s}\).

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