A shell is fired from a point O at an angle of `60^(@)` with a speed of `40 m//s` & strikes a horizontal plane throught O, at a point A. The gun is fired a second time with the same angle of elevation but a different speed v. If it hits the target which starts to rise vertically from A with a constant speed `9sqrt(3) m//s` at the same instant as the shell is fired, find v.

Explanation

To solve the problem, we need to analyze the motion of the shell fired from point O and the target that starts rising from point A. Here’s a step-by-step solution:

### Step 1: Determine the time of flight of the first shell
The shell is fired at an angle of \(60^\circ\) with an initial speed of \(40 \, \text{m/s}\). The time of flight \(T\) for a projectile is given by the formula:

\[ T = \frac{2u \sin \theta}{g} \]

where:
- \(u = 40 \, \text{m/s}\) (initial speed),
- \(\theta = 60^\circ\),
- \(g = 9.81 \, \text{m/s}^2\) (acceleration due to gravity).

Calculating \(T\):

\[ T = \frac{2 \times 40 \times \sin(60^\circ)}{9.81} \]
\[ = \frac{2 \times 40 \times \frac{\sqrt{3}}{2}}{9.81} \]
\[ = \frac{40\sqrt{3}}{9.81} \]

### Step 2: Calculate the height of the shell at time \(T\)
The maximum height \(H\) reached by the shell can be calculated using:

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

Calculating \(H\):

\[ H = \frac{40^2 \sin^2(60^\circ)}{2 \times 9.81} \]
\[ = \frac{1600 \times \frac{3}{4}}{19.62} \]
\[ = \frac{1200}{19.62} \]

### Step 3: Determine the height of the target at time \(T\)
The target rises vertically from point A with a constant speed of \(9\sqrt{3} \, \text{m/s}\). The height \(h\) of the target after time \(T\) is given by:

\[ h = v_t \cdot T \]

where \(v_t = 9\sqrt{3} \, \text{m/s}\).

Calculating \(h\):

\[ h = 9\sqrt{3} \cdot \frac{40\sqrt{3}}{9.81} \]
\[ = \frac{360}{9.81} \]

### Step 4: Set the heights equal to find \(v\)
For the second shell fired at speed \(v\) to hit the target, the height of the second shell must equal the height of the target at time \(T\).

Using the same formula for maximum height for the second shell:

\[ H' = \frac{v^2 \sin^2(60^\circ)}{2g} \]

Setting \(H' = h\):

\[ \frac{v^2 \cdot \frac{3}{4}}{2 \cdot 9.81} = \frac{360}{9.81} \]

### Step 5: Solve for \(v\)
Canceling \(9.81\) from both sides and simplifying gives:

\[ \frac{v^2 \cdot \frac{3}{4}}{2} = 360 \]
\[ v^2 \cdot \frac{3}{8} = 360 \]
\[ v^2 = 360 \cdot \frac{8}{3} \]
\[ v^2 = 960 \]
\[ v = \sqrt{960} = 4\sqrt{60} \approx 39.2 \, \text{m/s} \]

### Final Answer
The speed \(v\) of the second shell is approximately \(39.2 \, \text{m/s}\).

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