An artillery shell is launched on a flat, horizontal field at an angle of α = 31.5° with respect to the horizontal and with an initial speed of v0 = 266 m/s.
What is the horizontal distance covered by the shell after 9.44 s of flight?
What is the height of the shell at this moment?
To find the horizontal distance covered by the shell after 9.44 s of flight, we can use the equation for horizontal distance:
Horizontal distance = Initial horizontal velocity × time
To find the initial horizontal velocity, we can use the equation:
Initial horizontal velocity = Initial velocity × cos(α)
Substituting the given values:
Initial horizontal velocity = 266 m/s × cos(31.5°)
Now that we have the value for the initial horizontal velocity, we can calculate the horizontal distance:
Horizontal distance = Initial horizontal velocity × time
To find the height of the shell at this moment, we can use the equation for vertical displacement:
Vertical displacement = Initial vertical velocity × time + (1/2) × acceleration due to gravity × time squared
To find the initial vertical velocity, we can use the equation:
Initial vertical velocity = Initial velocity × sin(α)
Substituting the given values:
Initial vertical velocity = 266 m/s × sin(31.5°)
Now that we have the value for the initial vertical velocity, we can calculate the vertical displacement:
Vertical displacement = Initial vertical velocity × time + (1/2) × acceleration due to gravity × time squared
Please note that the acceleration due to gravity is approximately 9.8 m/s^2.