A Mortar shell is fired with muzzle speed of 500ft/sec.Find the angle of elevation of mortar if the shell strikes a target located 1200ft away. What is the maximum height of the shell?
a. Range = Vo^2*sin(2A)/g
Range = 1200 Ft.
Vo = 500 Ft./s.
g = 32 Ft/s^2
Solve for A.
b. V^2 = Vo^2 + 2g*h
h = -Vo^2/2g
Vo = 500 Ft./s.
g = -32 Ft./s^2
Solve for h
To find the angle of elevation of the mortar, as well as the maximum height of the shell, we can use principles of projectile motion. Here's how:
1. Determine the initial velocity: Given that the muzzle speed of the mortar shell is 500 ft/sec, the initial velocity of the shell is 500 ft/sec.
2. Find the time of flight: The time it takes for the shell to reach the target can be calculated using the formula: time = distance / velocity. In this case, the distance is 1200 ft, and the velocity is the horizontal component of the initial velocity, which can be found using the equation: velocity_horizontal = velocity * cos(theta), where theta is the angle of elevation. Substituting the given values, we get: time = 1200 ft / (500 ft/sec * cos(theta)).
3. Calculate the maximum height: The maximum height reached by the mortar shell can be found using the formula: height = (velocity * sin(theta))² / (2 * g), where g is the acceleration due to gravity (approximately 32.2 ft/sec²). By substituting the given values, we get: height = (500 ft/sec * sin(theta))² / (2 * 32.2 ft/sec²).
4. Solve for the angle of elevation: Since we have two unknowns, the angle of elevation (theta) and the time of flight, we can use the maximum height equation calculated in step 3 to create a relationship between theta and the time of flight. Rearranging the equation, we get: sin(theta) = √((2 * height * g) / (velocity²)). Now, substitute the known values, and you will get an equation that can be solved for theta.
By following these steps, you can find the angle of elevation of the mortar and the maximum height of the shell.
To find the angle of elevation and the maximum height of the shell, we can use the equations of projectile motion.
Let's assume that the angle of elevation is θ.
1. Finding the time of flight:
The horizontal distance traveled by the shell can be calculated using the equation:
Range = Velocity × Time
1200 ft = (500 ft/sec) × Time
Solving for Time, we have:
Time = 1200 ft / 500 ft/sec
Time ≈ 2.4 seconds
2. Finding the vertical component of velocity:
The vertical component of velocity can be calculated using the equation:
Vertical component of velocity = Velocity × sin(θ)
500 ft/sec × sin(θ) = Vertical component of velocity
3. Finding the maximum height:
Using the formula for the maximum height reached by a projectile, which is given by:
Max height = (Vertical component of velocity)^2 / (2 × gravitational acceleration)
Max height = (Vertical component of velocity) × (Vertical component of velocity) / (2 × 32.2 ft/sec^2)
Max height = (500 ft/sec × sin(θ)) × (500 ft/sec × sin(θ)) / (2 × 32.2 ft/sec^2)
Max height ≈ (sin(θ))^2 × 7.78 ft
To find the angle of elevation, we can use trigonometry:
sin(θ) = Opposite / Hypotenuse
sin(θ) = Max height / Range
sin(θ) = (sin(θ))^2 × 7.78 ft / 1200 ft
Simplifying the equation:
1 = (sin(θ)) × 7.78 ft / 1200 ft
Solving for sin(θ):
sin(θ) ≈ 1200 ft / (7.78 ft × 1)
sin(θ) ≈ 0.154
To find the angle of elevation, we take the inverse sine of 0.154:
θ ≈ arcsin(0.154)
θ ≈ 8.9 degrees
Therefore, the angle of elevation of the mortar is approximately 8.9 degrees. The maximum height of the shell is approximately (sin(θ))^2 × 7.78 ft, which is approximately 0.0198 * 7.78 ft ≈ 0.154 ft.