A man in a balloon, rising vertically with an acceleration of `5 m//s^(2)`, releases a ball 10 s after the balloon is let go from the ground. The greatest height above the ground reached by the ball is

Options

A. (a)`14.7 m`
B. (b)`19.6 m`
C. (c)` 9.8 m`
D. (d)` 24.5m `

Explanation

To solve the problem step by step, we will follow the physics concepts of kinematics.

### Step 1: Determine the velocity of the balloon at the time the ball is released.
The balloon rises with an acceleration \( a = 5 \, \text{m/s}^2 \) and starts from rest, so the initial velocity \( u = 0 \). The time \( t \) when the ball is released is 10 seconds.

Using the formula for velocity:
\[ v = u + at \]
Substituting the values:
\[ v = 0 + (5 \, \text{m/s}^2)(10 \, \text{s}) = 50 \, \text{m/s} \]

### Step 2: Calculate the height of the balloon when the ball is released.
We can use the formula for displacement (height in this case):
\[ h = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ h = 0 \cdot 10 + \frac{1}{2} \cdot 5 \cdot (10)^2 = \frac{1}{2} \cdot 5 \cdot 100 = 250 \, \text{m} \]

### Step 3: Determine the additional height the ball reaches after being released.
When the ball is released, it has an initial upward velocity of \( 50 \, \text{m/s} \) and will rise until its velocity becomes zero under the influence of gravity. The acceleration due to gravity \( g \) is approximately \( 9.8 \, \text{m/s}^2 \) (we can use \( 10 \, \text{m/s}^2 \) for simplicity).

Using the formula:
\[ v^2 = u^2 + 2as \]
Where \( v = 0 \) (final velocity), \( u = 50 \, \text{m/s} \) (initial velocity), and \( a = -g = -10 \, \text{m/s}^2 \):
\[ 0 = (50)^2 + 2(-10)s \]
\[ 0 = 2500 - 20s \]
\[ 20s = 2500 \implies s = \frac{2500}{20} = 125 \, \text{m} \]

### Step 4: Calculate the total height above the ground reached by the ball.
The total height \( H \) is the sum of the height of the balloon when the ball was released and the additional height the ball rises after being released:
\[ H = h + s = 250 \, \text{m} + 125 \, \text{m} = 375 \, \text{m} \]

### Final Answer:
The greatest height above the ground reached by the ball is \( 375 \, \text{m} \).

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