A ball is thrown vertically upwards with a velocity of 10 `ms^(-1)`. It returns to the ground with a velocity of 9 `ms^(-1)`. If g=9.8 `ms^(-2)` , then the maximum height attained by the ball is nearly (assume air resistance to be uniform)

Options

A. 5.1 m
B. 4.1 m
C. 4.61 m
D. 5 m

Explanation

To solve the problem step by step, we will analyze the motion of the ball when it is thrown upwards and when it returns downwards, taking into account the effects of air resistance.

### Step 1: Understand the motion of the ball
When the ball is thrown upwards, it has an initial velocity \( u = 10 \, \text{m/s} \). As it rises, it decelerates due to gravity \( g = 9.8 \, \text{m/s}^2 \) and air resistance \( R \). At the maximum height, its final velocity \( v = 0 \, \text{m/s} \).

### Step 2: Write the equation of motion for upward motion
Using the equation of motion:
\[ v^2 = u^2 - 2a_1h \]
where \( a_1 = g + \frac{R}{m} \) (since both gravity and air resistance act downwards), we can substitute the values:
\[ 0 = (10)^2 - 2(g + \frac{R}{m})h \]
This simplifies to:
\[ 100 = 2(g + \frac{R}{m})h \quad \text{(Equation 1)} \]

### Step 3: Write the equation of motion for downward motion
When the ball falls back down, it has a final velocity \( v = 9 \, \text{m/s} \) and an initial velocity \( u = 0 \, \text{m/s} \). The acceleration \( a_2 = g - \frac{R}{m} \) (gravity acts downwards while air resistance acts upwards). Using the equation of motion:
\[ v^2 = u^2 + 2a_2h \]
we substitute the values:
\[ (9)^2 = 0 + 2(g - \frac{R}{m})h \]
This simplifies to:
\[ 81 = 2(g - \frac{R}{m})h \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( 100 = 2(g + \frac{R}{m})h \)
2. \( 81 = 2(g - \frac{R}{m})h \)

We can add these two equations to eliminate \( R \):
\[ 100 + 81 = 2g h + 2g h \]
\[ 181 = 4gh \]
\[ h = \frac{181}{4g} \]

### Step 5: Substitute the value of \( g \)
Substituting \( g = 9.8 \, \text{m/s}^2 \):
\[ h = \frac{181}{4 \times 9.8} = \frac{181}{39.2} \approx 4.61 \, \text{m} \]

### Final Answer
The maximum height attained by the ball is approximately \( 4.61 \, \text{m} \).

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