"Projectile Motion Projectile Motion Projectile motion a combination of"
Projectile Motion Projectile motion: a combination of horizontal motion with constant horizontal velocity and vertical motion with a constant downward acceleration due to gravity. Projectile motion refers to the motion of an object that is thrown, or projected, into the air at an angle. We restrict ourselves to objects thrown near the Earth’s surface as the distance traveled and the maximum height above the Earth are small compared to the Earth’s radius so that gravity can be considered to be constant. Projectile Motion The motion of a projectile is determined only by the object’s initial velocity and gravity. The vertical motion of a projected object is independent of its horizontal motion. The vertical motion of a projectile is nothing more than free fall. The one common variable between the horizontal and vertical motions is time. Path of a Projectile A projectile moves horizontally with constant velocity while being accelerated vertically. A right angle exists between the direction of the horizontal and vertical motion; the resultant motion in these two dimensions is a curved path. The path of a projectile is called its trajectory. The trajectory of a projectile in free fall is a parabola. Path of a Projectile Path of a Projectile vo = initial velocity or resultant velocity vx = horizontal velocity vyi = initial vertical velocity vyf = final vertical velocity R= maximum horizontal distance (range) x = horizontal distance Dy = change in vertical position yi = initial vertical position yf = final vertical position q = angle of projection (launch angle) H = maximum height ag = gravity = 9.8 m/s2 Path of a Projectile Path of a Projectile The horizontal distance traveled by a projectile is determined by the horizontal velocity and the time the projectile remains in the air. The time the projectile remains in the air is dependent upon gravity. Immediately after release of the projectile, the force of gravity begins to accelerate the projectile vertically towards the Earth’s center of gravity. Path of a Projectile The velocity vector vo changes with time in both magnitude and direction. This change is the result of acceleration in the negative y direction (due to gravity). The horizontal component (x component) of the velocity vo remains constant over time because there is no acceleration along the horizontal direction The vertical component (vy) of the velocity vo is zero at the peak of the trajectory. However, there is a horizontal component of velocity, vx, at the peak of the trajectory. Path of a Projectile Path of a Projectile In the prior diagram, r is the position vector of the projectile. The position vector has x and y components and is the hypotenuse of the right triangle formed when the x and y components are plotted. The velocity vector vot would be the displacement of the projectile if gravity were not acting on the projectile. The vector 0.5agt is the vertical 2 displacement of the projectile due to the downward acceleration of gravity. Together, this determines the vertical position for the projectile: Δy = (vy·t) – (0.5·ag·t2) Path of a Projectile Problem Solving: Projectile Motion Analyze the horizontal motion and the vertical motion separately. If you are given the velocity of projection, vo, you may want to resolve it into its x and y components. Think for a minute before jumping into the equations. Remember that vx remains constant throughout the trajectory, and that vy = 0 m/s at the highest point of any trajectory that returns downward. Horizontal velocity component: vx is constant because there is no acceleration in the horizontal direction if air resistance is ignored. v x vo cos q Vertical velocity component: At the time of launch: v y i v o sin θ After the launch: v y f v o sinθ a g t If vy positive, direction of vertical motion is up; if vy negative, direction of vertical motion is down; if vy = 0, projectile is at highest point. Horizontal position component: x vx t x v o cos θ t If you launch the projectile horizontally: then vo = vx vyi = 0 m/s q = 0o Vertical position component: Δy y f y i y f y i v yi t 0.5 a g t 2 y f y i v o sin θ t 0.5 a g t 2 Relationship Between Vertical and Horizontal Position: x ag 2 Δy x tan θ 2 v 2 cos θ2 o this equation is only valid for launch angles in the range 0 < q < 90 Range (total horizontal displacement) v o sin 2 θ 2 R ag double angle formula : sin (2 θ) 2 sin θ cos θ v o 2 sin θ cos θ 2 R ag Maximum Height v o sin θ 2 2 H 2ag When Do The Range & Maximum Height Equations Work? Works when Dy = 0. Does not work when Dy 0. Determining vo from vx and vy If the vertical and horizontal components of the velocity are known, then the magnitude and direction of the resultant velocity can be determined. Magnitude: vo v x v y 2 2 Determining vo from vx and vy Direction: from the horizontal vy θ tan 1 v x Direction: from the vertical vx 1 θ tan v y Range and Angle of Projection Range and Angle of Projection The range is a maximum at 45 because sin (2·45) = 1. For any angle q other than 45, a point having coordinates (x,0) can be reached by using either one of two complimentary angles for q, such as 15 and 75 or 30 and 50 . Range and Angle of Projection The maximum height and time of flight differ for the two trajectories having the same coordinates (x, 0). A launch angle of 90° (straight up) will result in the maximum height any projectile can reach. For Objects Shot Horizontally: vx constant x m vx v yi 0 t s D x vx t Dy negative Dy = -height Δy 0.5 a g t 2 For Objects Shot Horizontally: When hits at bottom: Vyf should be negative vo = resultant velocity v y f a g t vo v x v y f 2 2 For Objects Shot Horizontally: q with horizontal: v yf 1 q tan vx q with vertical: vx 1 q tan v yf For Situations in Which Dy = 0 m v o 2 sin 2 θ R ag v o 2 sinθ 2 H 2ag v y i v o sinθ v x v o cosθ R vx t Δy v y i t 0.5 a g t 2 For Situations In Which Dy Positive For Situations In Which Dy Positive v x v o cos q v yi v o sin q At any point in the flight: Δy v y i t 0.5 a g t 2 x2 ag Δy x tanθ 2 v 2 cosθ 2 o vyf vyi ag t x vx t v x constant vo v x 2 v y f2 For Situations In Which Dy Negative For Situations In Which Dy Negative At launch: v x v o cos q v yi v o sin q After launch: x v x t Δy v y i t 0.5 a g t 2 x2 ag Δy x tanθ 2 v 2 cosθ 2 o v x constant When it hits ground: vyf vyi ag t vo v x v y f 2 2