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 step by step, we will follow the physics principles related to motion in a straight line, particularly focusing on projectile motion.

### Step 1: Determine the initial velocity (u)
We know that the maximum height (h) attained by the ball is 45 m. We can use the formula for maximum height in projectile motion:

\[ h_{max} = \frac{u^2}{2g} \]

Where:
- \( h_{max} = 45 \, \text{m} \)
- \( g = 10 \, \text{m/s}^2 \)

Substituting the values into the equation:

\[ 45 = \frac{u^2}{2 \times 10} \]

This simplifies to:

\[ 45 = \frac{u^2}{20} \]

Multiplying both sides by 20 to isolate \( u^2 \):

\[ u^2 = 45 \times 20 = 900 \]

Taking the square root to find \( u \):

\[ u = \sqrt{900} = 30 \, \text{m/s} \]

### Step 2: Set up the equation for velocity
According to the question, we need to find the time when the velocity of the ball becomes half of the initial velocity. Therefore, we need to find when:

\[ v = \frac{u}{2} = \frac{30}{2} = 15 \, \text{m/s} \]

### Step 3: Use the equation of motion to find time (t)
The velocity of the ball at any time \( t \) can be expressed as:

\[ v = u - gt \]

Substituting the known values into the equation:

\[ 15 = 30 - 10t \]

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

\[ 10t = 30 - 15 \]
\[ 10t = 15 \]
\[ t = \frac{15}{10} = 1.5 \, \text{s} \]

### 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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