a trench mortar fires a shell at an angle of 30 degrees with the horizontal and with a speed of 500 m/s. find its range and the maximum height it attains?
Range = (Vo^2/g)*sin(2*30)
= 22,092 meters
Max height H:
g H = (Vo*sin30)^2/2 = Vo^2/8
H = Vo^2/(8 g) = 3189 m
To find the range and maximum height attained by the trench mortar shell, we can use the equations of projectile motion.
1. Range:
The range of a projectile can be calculated using the horizontal component of the initial velocity (u) and the time of flight (t). The horizontal component of the initial velocity is given by u_x = u * cos(theta), where u is the initial velocity and theta is the launch angle.
Given:
u = 500 m/s
theta = 30 degrees
First, we calculate the horizontal component of the initial velocity:
u_x = 500 m/s * cos(30 degrees) = 500 m/s * 0.866 = 433 m/s
Next, we calculate the time of flight. The time of flight of a projectile launched at an angle can be calculated using the formula:
t = 2 * u * sin(theta) / g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values:
t = 2 * 500 m/s * sin(30 degrees) / 9.8 m/s^2 = 1000 m/s * 0.5 / 9.8 m/s^2 = 51.02 s (approx.)
Finally, we calculate the range as:
Range = u_x * t = 433 m/s * 51.02 s = 22,093.66 m (approx.)
Therefore, the range of the shell is approximately 22,094 meters.
2. Maximum Height:
The maximum height attained by the projectile can be calculated using the vertical component of the initial velocity (u_y) and the time of flight (t).
The vertical component of the initial velocity is given by u_y = u * sin(theta).
Plugging in the values:
u_y = 500 m/s * sin(30 degrees) = 500 m/s * 0.5 = 250 m/s
The maximum height (H) can be calculated using the formula:
H = u_y^2 / (2 * g)
Plugging in the values:
H = (250 m/s)^2 / (2 * 9.8 m/s^2) = 62500 m^2/s^2 / 19.6 m/s^2 ≈ 3193.88 m (approx.)
Therefore, the maximum height attained by the shell is approximately 3194 meters.
To find the range and maximum height attained by the trench mortar shell, we can use the equations of projectile motion.
Let's break down the given information:
- Launch angle (θ) = 30 degrees
- Initial velocity (v) = 500 m/s
First, we need to find the horizontal and vertical components of the initial velocity.
Horizontal component (v_x) = v * cos(θ)
v_x = 500 m/s * cos(30°)
v_x ≈ 500 * 0.866
v_x ≈ 433 m/s
Vertical component (v_y) = v * sin(θ)
v_y = 500 m/s * sin(30°)
v_y ≈ 500 * 0.5
v_y ≈ 250 m/s
Now, we can calculate the time it takes for the shell to reach the maximum height.
v_y = u_y + a * t
0 = 250 m/s - 9.8 m/s^2 * t_max_height
t_max_height ≈ 250 m/s / 9.8 m/s^2
t_max_height ≈ 25.51 seconds
Since the trajectory of the projectile is symmetrical, the time taken to reach the maximum height is the same as the time taken to fall back to the ground. Therefore, the total time of flight is double the time to reach the maximum height:
Total time of flight (t_total) = 2 * t_max_height
t_total ≈ 2 * 25.51 seconds
t_total ≈ 51.02 seconds
Next, we can find the total horizontal distance (range) covered by the shell.
Range (R) = v_x * t_total
R ≈ 433 m/s * 51.02 seconds
R ≈ 22,066.66 meters
Therefore, the range of the trench mortar shell is approximately 22,067 meters.
Finally, we can determine the maximum height (H) attained by the shell.
H = v_y^2 / (2 * g)
H ≈ (250 m/s)^2 / (2 * 9.8 m/s^2)
H ≈ 62,500 m^2/s^2 / 19.6 m/s^2
H ≈ 3,191.32 meters
Hence, the maximum height attained by the trench mortar shell is approximately 3,191 meters.