A ball is thrown vertically upward attains a maximum height of 45 m. The time after which velocity of the ball become equal to half the velocity of projection ? (use g = 10 `m//s^(2)`)

Options

A. 2 s
B. 1.5 s
C. 1 s
D. 0.5 s

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the initial velocity (u)
We know that the maximum height (h) attained by the ball is given by the formula:
\[ h = \frac{u^2}{2g} \]
Where:
- \( h = 45 \, \text{m} \)
- \( g = 10 \, \text{m/s}^2 \)

Substituting the values into the equation:
\[ 45 = \frac{u^2}{2 \times 10} \]
\[ 45 = \frac{u^2}{20} \]
Multiplying both sides by 20:
\[ u^2 = 45 \times 20 \]
\[ u^2 = 900 \]
Taking the square root of both sides:
\[ u = \sqrt{900} \]
\[ u = 30 \, \text{m/s} \]

### Step 2: Determine the final velocity (v)
We need to find the time after which the velocity of the ball becomes half of the initial velocity. Therefore, the final velocity (v) is:
\[ v = \frac{u}{2} = \frac{30}{2} = 15 \, \text{m/s} \]

### Step 3: Use the second equation of motion
We will use the second equation of motion:
\[ v = u + at \]
Where:
- \( v = 15 \, \text{m/s} \) (final velocity)
- \( u = 30 \, \text{m/s} \) (initial velocity)
- \( a = -g = -10 \, \text{m/s}^2 \) (acceleration due to gravity, negative because it acts downward)
- \( t \) is the time we need to find.

Substituting the values into the equation:
\[ 15 = 30 - 10t \]

### Step 4: Solve for time (t)
Rearranging the equation:
\[ 10t = 30 - 15 \]
\[ 10t = 15 \]
Dividing both sides by 10:
\[ t = \frac{15}{10} \]
\[ t = 1.5 \, \text{s} \]

### Final Answer
The time after which the velocity of the ball becomes equal to half the velocity of projection is **1.5 seconds**.

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