Kinematics Projectile Motion Objectives Define a projectile Describe

Kinematics Projectile Motion

Objectives • Define a projectile • Describe the trajectory • State which variables remain constant during a projectile motion • Solve projectile motion problems • Sketch the kinematic graphs for projectile motion

Projectile An object launched, thrown, or fired under the influence of gravity only. The trajectory of a projectile is parabolic What is the significance of “gravity only” ? Gravity refers to the acceleration of gravity. “Gravity only” implies that the acceleration of gravity is the only acceleration that projectiles experience. The acceleration of gravity operates in the vertical direction only, which means that projectiles are accelerated in the y-direction. There is NO horizontal acceleration due to gravity, which means that projectiles move at constant velocity in the x-direction.

Components of the Initial Launch Velocity In most problems the launch velocity is given along with the angle of the launch. The angle of the initial launch is usually in the 1 st or 4 th quadrants. The first step in most problems is determining the x- and y-components of the initial velocity vector. Examples: 50 m/s, 37 o above +x 50 m/s, 37 o below +x 50 m/s horizontally v 0 x = v 0 cos v 0 x = 50 cos 37 o v 0 x = +40 m/s v 0 x = v 0 cos v 0 x = 50 cos 0 o v 0 x = +50 m/s v 0 y = v 0 sin v 0 y = 50 sin 37 o v 0 y = +30 m/s v 0 y = v 0 sin v 0 y = 50 sin 37 o v 0 y = 30 m/s v 0 y = v 0 sin v 0 y = 50 sin 0 o v 0 y = 0 m/s

Independence of Motion Projectiles accelerate in the y-direction and move at constant velocity in the x-direction. The mathematics of these two motions is completely different. As before the motion in the x- and y-directions are mathematically independent. You can solve in the x-direction using only x-variables and the constant velocity equation, while ignoring the y-direction completely. You can also solve the y-direction using only y-variables and the kinematic equations for acceleration, while ignoring the x-direction. While the x- and y-motions solve independently they take place in simultaneous time. Time is the one variable shared by both the x- and y-directions. Once time is solved in one direction, it can then be applied to solve equations in the other direction. This makes time a critical variable.

Projectile Motion Variables v 0 θ 0 v 0 y= v 0 sin θ v 0 x= v 0 cos θ Most projectile motion problems begin by giving launch data. v 0 Magnitude of initial launch velocity. θ 0 Angle of the launch velocity. When ever you encounter a vector at an angle you must find its components. This is a critical step in many problems through out the entire year, and should be done automatically from now on. v 0 x Component of the initial velocity in the x direction, v 0 x = v 0 cos θ v 0 y Component of the initial velocity in the y direction, v 0 y = v 0 sin θ

Projectile Motion Variables v 0 θ 0 v 0 y= v 0 sin θ v 0 x= v 0 cos θ Δy Δx The projectile then travels in a parabolic trajectory. It may experience displacements in one or both directions (x and/or y). Δx Range (change in horizontal position measured from the launch). Δy Altitude (change in vertical position measured from the launch). In a time t Time of flight

Projectile Motion Variables v 0 θ 0 vx θ v 0 y= v 0 sin θ v 0 x= v 0 cos θ Δy v vy Δx The projectile finishes the problem with an overall final velocity. v Magnitude of final velocity at the designated end of the problem. θ Angle of the final velocity. This is also a vector at an angle, and it splits into components vx Component of final x direction velocity. vy Component of final y direction velocity.

Projectile Motion Equations v 0 θ 0 vx θ v 0 y= v 0 sin θ v 0 x= v 0 cos θ Original Equation Δy v vy Δx Modified for y direction Modified for x direction REMEMBER: Motion in the x direction is constant velocity, ax = 0.

Projectile Motion Equations vx v 0 θ v 0 y= v 0 sin θ θ 0 v 0 x= v 0 cos θ Result y-direction: Free fall X-direction: Constant velocity Δy v vy Δx Modified for y direction Modified for x direction

B Sign on the Variables Identify the correct variable signs at key places along a projectiles trajectory. A C D E A Moving upward B At max height C Moving downward, above initial position D Moving downward, even with initial position E Moving downward, below initial position Δy + + + 0 − vy + 0 − − −

Analyzing an Upward Launch For this presentation g = 10 m/s 2. This simplifies example calculations. x-direction y-direction Split into x and y adding subscripts ax = 0 and ay = g Horizontal displacement due to the x velocity. Vertical displacement due to the y velocity. Vertical displacement due to gravity g.

100 The graph at the left is a position graph ( y – x ). 80 Plot a projectile that is launched at 28. 284 m/s at an angle of 45 o. Step 0: Find components. 60 y (m) 40 If there was no gravity the object would move at constant velocity in both the x and y directions. 20 v 0 0 − 20 − 40 v 0 x v 0 y 20 40 x (m) 60 80 100 What does gravity do? Gravity only affects the y direction adding acceleration to the equation. It would move 20 m right and 20 m up every second resulting in the path described by the dashed line.

t 100 0 80 1 60 y (m) 2 40 20 0 3 20 40 x (m) 60 80 100 4 − 20 5 − 40

100 80 t The dashed line represent the effect of velocity alone. 0 1 60 y (m) 2 40 20 0 3 20 40 x (m) 60 80 100 4 − 20 5 − 40

100 80 − 125 The drop due to gravity always follows this pattern. − 80 0 − 45 60 y (m) t 1 − 20 2 40 − 5 20 0 3 20 40 x (m) 60 80 100 4 − 20 5 − 40

Examine the Components of Velocity Each Second +10 m/s +20 m/s 20 +20 m/s y (m) − 10 m/s 0 +20 m/s 20 40 x (m) 60 +20 m/s 80 − 20 m/s x-direction is constant velocity. In the y-direction gravity takes away 10 m/s (9. 8) every second. 100 +20 m/s − 30 m/s

Facts Worth Knowing +10 m/s +20 m/s 20 +20 m/s y (m) − 10 m/s 0 +20 m/s 20 40 x (m) 60 +20 m/s 80 100 − 20 m/s x-motion is always constant velocity and positive. Any two point with the same altitude The speed (magnitude of velocity) is the same. The direction is opposite: Positive on the way up and negative on the way down At the very top there is no y-velocity Velocity at max height equals the initial x-direction velocity, vmax height = v 0 x.

Solving Max Height Problems Determining maximum height HIDDEN ZERO: At maximum height the vertical velocity is zero, vy = 0. While the projectile does have a horizontal velocity, this is irrelevant when solving independently in the y-direction. The best equation to determine maximum height is: Substituting zero for vy and rearranging for height results in: Determining the speed or velocity at maximum height This you should already know. The instantaneous velocity at max height is horizontal, and since velocity is constant in the x-direction the velocity and speed at max height are equal to the x-component of the initial velocity.

Example 1 A projectile is launched at an angle of 37 o above the horizontal with an initial speed of 50 m/s. When given the magnitude and direction of the initial launch determine the components of initial velocity first. v 0 θ 0 v 0 x vx v 0 y a. Determine the max height reached by the projectile. b. Determine the time to reach max height. c. Determine the speed of the projectile at max height. Δy Δx v θ v y

Solving For Locations Other Than Maximum Height Most problems begin by solving one of the following equations for time, and then substituting time into the other equation. The equation on the left is solved exactly the same as it was in the free fall assignment, and the equation on the right is pretty simple. Keep three things in mind: • Use the components of the initial velocity, and NOT the initial velocity. • The left equation can only contain y-direction variable • The right equation can only contain x-direction variable. Use the other equations as necessary.

Example 2 A projectile is launched at an angle of 37 o above the horizontal with an initial speed of 50 m/s. During its flight the projectile achieves a range of 200 m. v 0 x = 40 m/s , and v 0 y = 30 m/s a. Determine the time of flight. v 0 θ 0 v 0 x vx v 0 y Δy v θ v y Δx b. Determine the height of the projectile relative to the launch point. c. Determine the speed of the projectile. NOTE: This refers to the diagonal speed

Example 3 A projectile is launched at an angle of 37 o above the horizontal with an initial speed of 50 m/s. The projectile’s time of flight is 5 seconds. a. Determine the range of the projectile. v 0 θ 0 v 0 x vx v 0 y Δy v θ v y Δx b. Determine the height of the projectile relative to the launch point. c. Determine the speed of the projectile. NOTE: This refers to the diagonal speed The solution to this part solved the same as Example 1

Example 4 A projectile is launched at an angle of 37 o above the horizontal with an initial speed of 50 m/s. The projectile lands 25 m above its initial launch height. a. Determine the time of flight. v 0 θ 0 v 0 x vx v 0 y Δy v θ v y Δx b. Determine the projectiles range. c. Determine the speed of the projectile. NOTE: This refers to the diagonal speed See the two previous examples. This calculates the same each time.

Two Special Launches There are two frequently encountered launches, that have hidden zero quantities. Hidden zeros often means short cuts and easier solutions. Landing at the starting height Horizontal Launches Hidden zero: y = 0 Hidden zero: v 0 y = 0

Landing at the Initial Height 0 m/s +20 m/s vtop = v 0 x 20 +20 m/s v 0 y (m) θ 0 +20 m/s 20 40 x (m) 80 60 −θ 100 v = v 0 − 20 m/s Maximum Altitude Return to Initial Height Other Info 2 nd half look like a horizontal launch.

Landing at the Initial Height There is a pattern associated with launch angles for objects returning to their initial height. Any two angles of launch that add up to 90 o will arrive at the same landing position. The greatest range is achieved by launching at an angle of 45 o. Maximum height is achieved by launching at 90 o. 75 o 60 o 45 o 30 o 15 o

Example 5 A projectile is launched at 141. 42 m/s to achieve maximum range. When given launch velocity and launch angle, determine launch components. HIDDEN VALUE: Maximum range means the launch angle is 45 o. a. Determine the projectile’s maximum height. HIDDEN ZERO: At max height vy = 0. b. Determine the time to reach maximum height. c. Determine the projectile’s speed at maximum height.

Example 5 A projectile is launched at 141. 42 m/s to achieve maximum range. d. Determine the total time of flight to landing. HIDDEN ZERO: There is no mention of landing higher or lower, y = 0. e. Determine the projectile’s maximum range. This is 4 times the maximum height. Is this always true in max range problem? f. Determine the projectile’s landing speed.

Example 5 A ball is thrown horizontally with a speed of 25 m/s from the top of a 45 m tall building. When given launch velocity and launch angle, determine launch components. HIDDEN ZERO: In horizontal launches v 0 y = 0. a. Determine the time of flight. Just as in free fall problems, the vertical displacement is negative. This means the minus signs will cancel. Everyone knows this, and for dropped objects in free fall and horizontally launched projectiles, most people ignore the minus signs. b. Determine the projectile’s range.

Graphing a Horizontal Launch x (m) y (m) 0 20 40 0 t (s) 0 1 20 40 2 60 3 4 80 Clearly x and y solved independently, yet the events took place in simultaneous time 60 80

Projectile Motion Graphs Position Graphs: These look exactly like the flight path. y Horizontal Launch y x Upward Launch x Acceleration Graphs: There is only one acceleration, gravity = − 9. 8 g Horizontal Launch g t Upward Launch t

Projectile Motion Graphs Horizontal Velocity Graphs: Constant Velocity (no acceleration in x ) vx Horizontal Launch vx t Upward Launch t Vertical Velocity Graphs: a is the rate of Δv (subtract 9. 8 each second) vy Horizontal Launch Starts at rest vertically vy t Upward Launch Thrown up (+vy) initially t