without air resistance maximum range of projectile is obtained with an angle of ????
Can you please fill in the blank THANK YOU
at an angle of 45 degrees
45 degrees
To determine the angle at which a projectile achieves maximum range without air resistance, we can use the concept of projectile motion.
The equation for the range (horizontal distance traveled) of a projectile without air resistance is:
Range = (v^2 * sin(2θ)) / g
Where:
- v is the initial velocity of the projectile
- θ is the launch angle (the angle at which the projectile is projected)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
To find the angle that maximizes the range, we need to maximize the expression sin(2θ).
When sin(2θ) is at its maximum value, it means that the angle (2θ) is equal to 90 degrees, because sin(90) = 1. Therefore, 2θ = 90 degrees, and θ = 45 degrees.
So, the angle at which a projectile without air resistance achieves maximum range is 45 degrees.
Certainly! To find the angle that gives the maximum range of a projectile without air resistance, we can use the concept of projectile motion. The maximum range is obtained when the projectile follows a parabolic path, reaching the same vertical height at both the launch and landing points.
The formula for the range of a projectile without air resistance is:
Range = (v^2 * sin(2θ))/g,
where:
- v is the initial velocity of the projectile,
- θ is the launch angle, and
- g is the acceleration due to gravity.
To find the launch angle that gives the maximum range, we need to maximize the sin(2θ) term. The maximum value of sin(2θ) is 1, which occurs when 2θ = 90 degrees. Therefore, the angle that gives the maximum range is:
2θ = 90 degrees,
θ = 45 degrees.
So the blank can be filled with "45 degrees."