An arrow is shot into air. Its range is 200m and its time of flight is 5s. If `g = 10 ms^(-2)`. 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 can break it down into two main parts: finding the horizontal component of the velocity and then calculating the maximum height.

### Step 1: Finding the Horizontal Component of Velocity

1. **Use the Range Formula**: The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

2. **Rearranging the Range Formula**: We can express the range in terms of the horizontal component of the velocity \( u \cos \theta \):
\[ R = 2u \sin \theta \cdot u \cos \theta \]
This can be rearranged to find \( u \cos \theta \):
\[ u \cos \theta = \frac{R \cdot g}{2u \sin \theta} \]

3. **Use the Time of Flight Formula**: The time of flight \( T \) is given by:
\[ T = \frac{2u \sin \theta}{g} \]
Rearranging gives:
\[ 2u \sin \theta = T \cdot g \]
Substituting the values \( T = 5 \, \text{s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ 2u \sin \theta = 5 \cdot 10 = 50 \, \text{m/s} \]

4. **Finding \( u \sin \theta \)**: We can find \( u \sin \theta \):
\[ u \sin \theta = \frac{50}{2} = 25 \, \text{m/s} \]

5. **Substituting into the Range Formula**: Now substituting \( u \sin \theta \) back into the range formula:
\[ R = 200 \, \text{m} \]
We can express \( u \cos \theta \) as:
\[ u \cos \theta = \frac{R \cdot g}{50} \]
Substituting the known values:
\[ u \cos \theta = \frac{200 \cdot 10}{50} = 40 \, \text{m/s} \]

### Step 2: Finding the Maximum Height

1. **Use the Maximum Height Formula**: The maximum height \( H \) is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

2. **Expressing in Terms of \( u \sin \theta \)**: We can rewrite this as:
\[ H = \frac{(u \sin \theta)^2}{2g} \]
Substituting \( u \sin \theta = 25 \, \text{m/s} \):
\[ H = \frac{25^2}{2 \cdot 10} = \frac{625}{20} = 31.25 \, \text{m} \]

### Final Answers

- The horizontal component of velocity \( u \cos \theta = 40 \, \text{m/s} \)
- The maximum height \( H = 31.25 \, \text{m} \)