If the horizontal range of a projectile be a and the maximum height attained by it is b, then prove that the velocity of projection is `[2g(b+a^2/(16b))]^(1/2)`

Options

A. `[g(b + (a^(2))/(16b))]^(1/2)`
B. `[2g(b + (a^(2))/(16b))]^(1/2)`
C. `[g(a + (b^(2))/(16b))]^(1/2)`
D. `[2g(a + (b^(2))/(16b))]^(1/2)`

Answer

To prove that the velocity of projection \( u \) is given by the formula

\[ u = \sqrt{2g\left(b + \frac{a^2}{16b}\right)}, \]

where \( a \) is the horizontal range and \( b \) is the maximum height, we can follow these steps:

### Step 1: Write the equations for range and height
For a projectile, the horizontal range \( R \) and the maximum height \( H \) are given by:

\[ R = \frac{u^2 \sin 2\theta}{g} \]
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

### Step 2: Express \( \sin 2\theta \) and \( \sin^2 \theta \)
Using the double angle identity, we have:

\[ \sin 2\theta = 2 \sin \theta \cos \theta \]

### Step 3: Divide the range equation by the height equation
Dividing the range equation by the height equation gives:

\[ \frac{R}{H} = \frac{u^2 \sin 2\theta / g}{u^2 \sin^2 \theta / (2g)} = \frac{2 \sin 2\theta}{\sin^2 \theta} \]

### Step 4: Substitute \( \sin 2\theta \)
Substituting \( \sin 2\theta \):

\[ \frac{R}{H} = \frac{2 \cdot 2 \sin \theta \cos \theta}{\sin^2 \theta} = \frac{4 \cos \theta}{\sin \theta} = 4 \cot \theta \]

### Step 5: Express \( \tan \theta \)
From the above, we can express \( \tan \theta \):

\[ \frac{R}{H} = 4 \cot \theta \implies \tan \theta = \frac{4H}{R} \]

### Step 6: Use Pythagorean theorem to find \( \sin \theta \)
In a right triangle where the opposite side is \( 4b \) and the adjacent side is \( a \):

\[ \text{Hypotenuse} = \sqrt{(4b)^2 + a^2} = \sqrt{16b^2 + a^2} \]

Thus,

\[ \sin \theta = \frac{4b}{\sqrt{16b^2 + a^2}} \]

### Step 7: Substitute \( \sin \theta \) into the height equation
Now substitute \( \sin \theta \) into the height equation:

\[ H = \frac{u^2 \sin^2 \theta}{2g} \implies b = \frac{u^2}{2g} \cdot \left(\frac{4b}{\sqrt{16b^2 + a^2}}\right)^2 \]

### Step 8: Simplify the equation
This gives:

\[ b = \frac{u^2 \cdot 16b^2}{2g(16b^2 + a^2)} \]

Rearranging gives:

\[ 2gb(16b^2 + a^2) = 16b^2u^2 \]

### Step 9: Solve for \( u^2 \)
This simplifies to:

\[ u^2 = \frac{2g(16b^2 + a^2)}{16b} \]

### Step 10: Factor out common terms
Factoring out gives:

\[ u^2 = g \left(\frac{16b^2}{8b} + \frac{a^2}{8b}\right) = g \left(2b + \frac{a^2}{16b}\right) \]

### Step 11: Final expression for \( u \)
Thus, we arrive at:

\[ u = \sqrt{2g\left(b + \frac{a^2}{16b}\right)} \]

### Conclusion
We have proved that the velocity of projection \( u \) is given by:

\[ u = \sqrt{2g\left(b + \frac{a^2}{16b}\right)} \]