An arrow is shot into air. Its range is 200m and its time of flight is 5s. If `g = 10ms^(-2)`, then horizontal component of velocity and the maximum height will be respectively

Options

A. `20 ms^(-1), 62.50 m`
B. `40 ms^(-1), 31.25m`
C. `80 ms^(-1), 62.5m`
D. None of these

Explanation

To solve the problem step by step, we will find the horizontal component of velocity and the maximum height of the arrow shot into the air.

### Step 1: Find the Horizontal Component of Velocity (Vx)

The formula for the range (R) of a projectile is given by:

\[ R = V_x \times T \]

Where:
- \( R \) = Range (200 m)
- \( V_x \) = Horizontal component of velocity
- \( T \) = Time of flight (5 s)

Rearranging the formula to find \( V_x \):

\[ V_x = \frac{R}{T} \]

Substituting the values:

\[ V_x = \frac{200 \, \text{m}}{5 \, \text{s}} = 40 \, \text{m/s} \]

### Step 2: Find the Vertical Component of Velocity (Vy)

The formula for the time of flight (T) is given by:

\[ T = \frac{2V_y}{g} \]

Where:
- \( V_y \) = Vertical component of velocity
- \( g \) = Acceleration due to gravity (10 m/s²)

Rearranging the formula to find \( V_y \):

\[ V_y = \frac{g \times T}{2} \]

Substituting the values:

\[ V_y = \frac{10 \, \text{m/s}^2 \times 5 \, \text{s}}{2} = 25 \, \text{m/s} \]

### Step 3: Find the Maximum Height (h)

The formula for maximum height (h) is given by:

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

Substituting the values we found for \( V_y \) and \( g \):

\[ h = \frac{(25 \, \text{m/s})^2}{2 \times 10 \, \text{m/s}^2} \]

Calculating:

\[ h = \frac{625 \, \text{m}^2/\text{s}^2}{20 \, \text{m/s}^2} = 31.25 \, \text{m} \]

### Final Answer

The horizontal component of velocity \( V_x \) is \( 40 \, \text{m/s} \) and the maximum height \( h \) is \( 31.25 \, \text{m} \).