a projectile is fired from the ground level with a velocity of 500ms at 30 degree to the horizontal. Determine 1 it's horizontal range 2 the greatest vertical height reached 3 the time taken to reach this height (g=10mls2
an anti aircraft gun fires at an elevation of 60degree at an enemy aircraft at 10.000 m above the ground. AE. what speed must the cannon be shot to hit the plane at the height (g=10m50)
To solve these projectile motion problems, we can use the basic principles of physics and the equations of motion. Here's how you can determine the horizontal range, greatest vertical height reached, and the time taken to reach this height:
1. Horizontal Range:
The horizontal range is the horizontal distance covered by the projectile.
Given:
- Initial velocity (u) = 500 m/s
- Launch angle (θ) = 30 degrees
- Acceleration due to gravity (g) = 10 m/s^2
The equation for horizontal range (R) is:
R = (u^2 * sin(2θ)) / g
Now, substitute the given values into the equation:
R = (500^2 * sin(2 * 30)) / 10
R = (250000 * sin(60)) / 10
R = (250000 * √3/2) / 10
R = 125000 * √3 / 20
R ≈ 1082.34 meters
Therefore, the horizontal range of the projectile is approximately 1082.34 meters.
2. Greatest Vertical Height Reached:
The greatest vertical height reached by the projectile is the maximum height above the ground level.
Using the equation for vertical displacement (h):
h = (u^2 * sin²(θ)) / (2g)
Substituting the given values:
h = (500^2 * sin²(30)) / (2 * 10)
h = (250000 * (1/2)^2) / 20
h = (250000 * 1/4) / 20
h = 62500 / 20
h = 3125 meters
Therefore, the greatest vertical height reached by the projectile is 3125 meters.
3. Time Taken to Reach Greatest Vertical Height:
To find the time taken to reach the greatest vertical height, we use the equation for time of flight (T):
T = (2 * u * sin(θ)) / g
Substituting the given values:
T = (2 * 500 * sin(30)) / 10
T = (1000 * 1/2) / 10
T = 500 / 10
T = 50 seconds
Therefore, the time taken to reach the greatest vertical height is 50 seconds.