A ball is thrown vertically upwards from the ground with a speed of `25.2 ms^(-1)`. How long does it take to reach its highest point and how high does it rise ? (Take `g=9.8 ms^(-2)`)

Options

A. `2.75s, 3.24 m`
B. `25.7s, 34.2 m`
C. `2.57s, 32.4 m`
D. `27.5s, 3.42 m`

Explanation

To solve the problem of a ball thrown vertically upwards with an initial speed of \( 25.2 \, \text{m/s} \) and to find out how long it takes to reach its highest point and how high it rises, we can follow these steps:

### Step 1: Identify the given values
- Initial speed (\( u \)) = \( 25.2 \, \text{m/s} \)
- Acceleration due to gravity (\( g \)) = \( 9.8 \, \text{m/s}^2 \)

### Step 2: Calculate the time taken to reach the highest point
At the highest point, the final velocity (\( v \)) of the ball will be \( 0 \, \text{m/s} \). We can use the formula:

\[ v = u - g \cdot t \]

Setting \( v = 0 \):

\[ 0 = 25.2 - 9.8 \cdot t \]

Rearranging the equation to solve for \( t \):

\[ 9.8 \cdot t = 25.2 \]

\[ t = \frac{25.2}{9.8} \]

Calculating \( t \):

\[ t \approx 2.57 \, \text{s} \]

### Step 3: Calculate the maximum height attained
We can use the formula for height (\( h \)) attained when an object is thrown upwards:

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

Substituting the values:

\[ h = \frac{(25.2)^2}{2 \cdot 9.8} \]

Calculating \( (25.2)^2 \):

\[ (25.2)^2 = 635.04 \]

Now substituting back into the height formula:

\[ h = \frac{635.04}{19.6} \]

Calculating \( h \):

\[ h \approx 32.4 \, \text{m} \]

### Final Answers
- Time taken to reach the highest point: \( t \approx 2.57 \, \text{s} \)
- Maximum height attained: \( h \approx 32.4 \, \text{m} \)

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