The minimum and maximum velocities of a projectile are 10`ms^(-1)` and 20`ms^(-1)` respectively. The horizontal range and maximum height are respectively (g=10 `ms^(-2)`)

Answer

To solve the problem, we need to find the horizontal range and maximum height of a projectile given its minimum and maximum velocities. Let's break it down step by step.

### Step 1: Understand the given data
- Minimum velocity (at maximum height), \( v_{\text{min}} = 10 \, \text{ms}^{-1} \)
- Maximum velocity (initial velocity), \( v_{\text{max}} = 20 \, \text{ms}^{-1} \)
- Acceleration due to gravity, \( g = 10 \, \text{ms}^{-2} \)

### Step 2: Relate the velocities to the angle of projection
At maximum height, the vertical component of the velocity becomes zero, and only the horizontal component remains. The horizontal component of the initial velocity \( u \) is given by:
\[ v_{\text{min}} = u \cos \theta \]
where \( u \) is the initial velocity and \( \theta \) is the angle of projection.

From the problem, we know:
\[ u = v_{\text{max}} = 20 \, \text{ms}^{-1} \]
Thus,
\[ 10 = 20 \cos \theta \]
Solving for \( \cos \theta \):
\[ \cos \theta = \frac{10}{20} = \frac{1}{2} \]
This gives:
\[ \theta = 60^\circ \]

### Step 3: Calculate the horizontal range
The formula for the horizontal range \( R \) of a projectile is:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Substituting the values:
\[ R = \frac{(20)^2 \sin(2 \times 60^\circ)}{10} \]
Calculating \( \sin(120^\circ) \):
\[ \sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Now substituting this back into the range formula:
\[ R = \frac{400 \cdot \frac{\sqrt{3}}{2}}{10} = \frac{400\sqrt{3}}{20} = 20\sqrt{3} \, \text{meters} \]

### Step 4: Calculate the maximum height
The formula for the maximum height \( H \) of a projectile is:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Substituting the values:
\[ H = \frac{(20)^2 \sin^2(60^\circ)}{2 \cdot 10} \]
Calculating \( \sin^2(60^\circ) \):
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \implies \sin^2(60^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
Now substituting this back into the height formula:
\[ H = \frac{400 \cdot \frac{3}{4}}{20} = \frac{300}{20} = 15 \, \text{meters} \]

### Final Answers
- Horizontal Range, \( R = 20\sqrt{3} \, \text{meters} \)
- Maximum Height, \( H = 15 \, \text{meters} \)

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