Describe how the range of a projectile varies with launch angle.
I will be happy to critique your thinking.
1--The study of projectile motion is made easy by breaking the initial velocity into its vertical and horizontal components. Thus, If a projectile is fired with an initial velocity of Vo at an angle "µ" to the horizontal, Vv = Vvertical = (Vo)sin(µ) and Vh = Vhorizontal = (Vo)cos(µ). The time of flight may be obtained by summing the rise time with the fall time. From Vf = Vo - gt and Vf = 0, Vv = gt making the rise time t1 = Vv/t. During this period of time, the projectile travels horizontally d = Vht1. During the rise time, t1, the projectile rises to a height of h = Vvt1 - g(t1^2)/2 which can now be written as h = g(t1^2) - g(t1^2)/2 or h = g(t1^2)/2. Clearly, the time, t2, required for the projectile to fall back to the ground derives from -h = -g(t2^2)/2 making t1 = t2 or the total time T = 2t1.
Combining these expressions, d = Vocos(µ)t = 2Vot1cos(µ) = 2Vo[Vosin(µ)/g]cos(µ) or
d = Vo^2(sin(2µ))/2g.
Eliminating t1 from Vosin(µ) = gt1 and h = gt1^2/2 yields the maximum height reached from
Vo^2(sin^2µ) = g^2(t1)^2 and h = gt1^2/2, t1^2 = Vo^2sin^µ/g^2 = 2h/g or
h = Vo^2(sin^2(µ))/2g.
2--Envision an airplane on a bombing run flying at a constant altitude "s" and with constant horizontal velocity "Vh". The objective target "O" is being observed by the bombardier. When should the bombardier release the bomb in order to hit the target? More specifically, at what angle to the vertical should the bombardier release the bomb?
Note that when the bomb is released, it retains the same forward velocity of the airplane. Therefore, the bomb travels horizontally with a constant velocity independently of the simulaneous accelerated motion toward the earth's surface. As a result, the bomb will strike a point on the ground well ahead of the airplane's position at the time of release. The horizontal distance ahead of the airplane, "x", will make an angle "ø" with the vertical. Having "s", the altitude of the airplane, "x", the horizontal distance from the local vertical to the target objective"O", and "t", the time from release to the target, we have s = gt^2/2, x = Vh(t) and x = s(tanø), combining these expressions to eliminate both "x" and "t", we derive tanø = Vhsqrt(2/gs). Therefore, the bombardier releases the bomb when his instruements indicate that the angle from his local vertical position to the target is
ø = arctan[Vh(sqrt(2/gs))].
(Note that the atmospheric friction effects are neglected in the above derivations.)
Atmospheric Drag Effects on Projectile Motion
Another factor that a projectile, or rocket, encounters during flight is the drag force on the vehicle produced by the dynamic pressure of the atmosphere on the nose of the vehicle and the friction drag developed along the total cylindrical length of the body.. This drag is directly proportional to the air density, the frontal area of the rocket, the square of the velocity and the coefficient of friction associated with the nose and body shape. The major contributor to the total drag force is usually the nose which steadily pierces the atmosphere at increasing velocity as the rocket climbs through, and possibly, out of the atmosphere. Even though the nose is usually cone shaped to reduce drag, it is this surface shape that is fighting its way through the air.
A secondary, though sometimes predominant, contributor to the total drag, is the friction drag produced along the entire length of the vehicle. The total drag force may be calculated from the equation D = Cd(d)AV^2/2 where Cd (see sub d) is the total integrated drag coefficient of the vehicle, nose plus body area, derived from wind tunnel tests, d is the mass density of the atmosphere, A is the frontal area of the rocket, and V is the velocity. When the projectile, or rocket, is first fired, or lifting off, the velocity is low but the air density is highest and as the velocity increases with increasing altitude, the air density decreases. However, the drag increases with the square of the velocity so has an overriding effect on the magnitude of the drag. For orbital launches, the air density ultimately drops to the point where air drag is no longer a concern.
The range of a projectile refers to the horizontal distance it covers from the point of launch to the point of landing. The range is influenced by various factors, including the launch angle. The launch angle is the angle at which the projectile is launched with respect to the horizontal.
To understand how the range of a projectile varies with launch angle, we need to consider the concept of projectile motion. Projectile motion is a form of motion in which an object or projectile is launched into the air and moves along a curved path under the influence of gravity.
When the launch angle of a projectile is increased from 0 degrees (horizontal) to 90 degrees (vertical), the range of the projectile follows a specific pattern:
1. 0 degrees (horizontal launch): When a projectile is launched horizontally, its initial velocity is purely horizontal. In this case, the range will be determined solely by the initial velocity and the time of flight. The higher the initial velocity, the greater the range.
2. Less than 45 degrees: As the launch angle increases but remains less than 45 degrees, the range increases. This is because at lower angles, the projectile has a significant horizontal velocity component in addition to the vertical component. This helps it cover more horizontal distance before hitting the ground.
3. 45 degrees: When the launch angle is exactly 45 degrees, the range is maximized for a given initial velocity. This is because at 45 degrees, the horizontal and vertical velocity components are equal, allowing the projectile to travel the greatest horizontal distance before landing.
4. Greater than 45 degrees: As the launch angle exceeds 45 degrees and approaches 90 degrees, the range starts to decrease. This is because as the angle increases, more of the initial velocity is directed vertically, resulting in a shorter horizontal distance.
At 90 degrees (vertical launch), the range becomes zero as the projectile goes straight up and falls back to the ground at the same point of launch.
In summary, the range of a projectile is maximum at a launch angle of 45 degrees and decreases as the angle increases or decreases from this optimal value.