A shell if fired from a point O at an angle of 60 degree with a speed of 40 m/s and it strikes a horizontal plane through 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 9root3 m/s at the same instant as the shell is fired, find v (take g = 10 m/s^2).

To solve this problem, we need to analyze the motion of the shell and the target separately. The shell is fired at an angle of 60 degrees with an initial speed of 40 m/s, and we need to determine the speed v of the second shell that will hit a target rising vertically at a constant speed of 9√3 m/s. Let's break this down step by step.

Step 1: Determine the time of flight for the first shell

The first shell is fired at an angle of 60 degrees with an initial speed of 40 m/s. We can calculate the time of flight using the formula:

• Time of flight (T) = (2 * u * sin(θ)) / g

Here, u is the initial speed (40 m/s), θ is the angle (60 degrees), and g is the acceleration due to gravity (10 m/s²).

Calculating the sine of 60 degrees:

• sin(60°) = √3 / 2

Now substituting the values into the formula:

• T = (2 * 40 * (√3 / 2)) / 10
• T = (80√3 / 20) = 4√3 seconds

Step 2: Calculate the horizontal range of the first shell

The horizontal range (R) can be calculated using the formula:

• Range (R) = u * cos(θ) * T

Calculating the cosine of 60 degrees:

• cos(60°) = 1 / 2

Now substituting the values:

• R = 40 * (1 / 2) * (4√3)
• R = 20 * 4√3 = 80√3 meters

Step 3: Determine the height of the target when the second shell is fired

Since the target starts rising at the same instant the second shell is fired, we need to find out how high the target has risen by the time the second shell reaches the horizontal distance of 80√3 meters.

The height (h) of the target after time T is given by:

• h = speed * time

Using the speed of the target (9√3 m/s) and the time of flight (4√3 seconds):

• h = 9√3 * 4√3 = 36 meters

Step 4: Calculate the required speed v for the second shell

Now, we need to find the speed v of the second shell that will hit the target at a height of 36 meters after traveling horizontally 80√3 meters. The time of flight for the second shell will be the same as the first shell, which is 4√3 seconds.

Using the vertical motion formula:

• h = u * sin(θ) * t - (1/2) * g * t²

Substituting the known values:

• 36 = v * (√3 / 2) * (4√3) - (1/2) * 10 * (4√3)²

Calculating the second term:

• (1/2) * 10 * (16 * 3) = 240

Now we can rewrite the equation:

• 36 = 2v√3 - 240

Rearranging gives:

• 2v√3 = 276
• v = 138 / √3
• v = 46√3 m/s

Final Result

The speed v of the second shell that will hit the target is 46√3 m/s.