Projectile Motion Physics level 1 Mr Creery y

Projectile Motion Physics level 1 – Mr. Creery y vs. x plane y vs. t plane x vs. t plane

Projectile Motion Terms • X velocity component – The velocity of just the x axis found by cos(theta)*v • Y velocity component - The velocity of just the y axis found by sin(theta)*v • Range - How far a projectile moves along the x-axis. • Height – How far a projectile moves along the y-axis • Time – Independent variable which is the same on both the x and the y axis.

Projectile Motion Concepts • Vertical (y) and Horizontal (x) components are independent of each other. • There is no acceleration on the horizontal (x) component. • The acceleration due to gravity is -9. 8 m/s 2 on the vertical (y) axis. • The vertical time is the same as the horizontal time. • Velocity at the top of the trajectory is 0 m/s and the acceleration is always the same.

Projectile Motion Charts Chart shows the horizontal and vertical velocity vectors. Chart show the vertical and horizontal constants. A projectile question chart.

Vertical and Horizontal Motion y vs. x plane y vs. t plane x vs. t plane

Projectiles Launched at an Angle

Projectiles Launched at an Angle 5 s 4 s 3 s 2 s 1 s 5 m 20 m 45 m 80 m 125 m • This diagram shows that there are specific vertical distances for the cannonball from the position of where it should be (with no gravity) to where it actually is (with gravity). • These distances are the SAME distances it would fall if it were dropped and were undergoing free fall for the given amount of time.

Projectile Motion Vertical Graphs for an object thrown upwards with an initial velocity vi . The object takes time to reach its maximum height after which it falls back to the ground. (a) position vs. time graph (b) velocity vs. time graph (c) acceleration vs. time graph.

Projectile Motion Horizontal Graphs

Projectile Motion Examples • Trajectories • For objects launched at an angle, you have to do a little more work to determine the initial velocity in both the horizontal and vertical directions. • For example, if a football is kicked with an initial velocity of 40 m/s at an angle of 30° above the horizontal. • First, break the initial velocity vector up into x- and y-components. • Then, use the components for initial velocities in your horizontal and vertical to solve for the time on the y and the range on the x. • Don’t forget that symmetry of motion also applies to the parabola of projectile motion. For objects launched and landing at the same height, the launch angle is equal to the landing angle. The launch velocity is equal to the landing velocity. And if you want an object to travel the maximum possible horizontal distance (or range), launch it at an angle of 45°.

Height • The vertical distance a projectile falls below an imaginary straight-line path increases continually with time and is equal to h=1/2 at 2 or 4. 9 t 2 meters.

Height • The diagram above shows the path of a projectile with velocity vectors drawn in for different points on its parabolic path. The vectors have been resolved into their x and y components.

Height • The horizontal component is always the same – the ball moves equal distances in equal time intervals. • Note that ONLY the vertical component changes – it decreases going upward against gravity and increases going downward with gravity. The projectile accelerates due to Earth’s gravity. • Is the velocity of the projectile at its highest point zero?

Range • Now suppose 2 projectiles are launched at the same speed but different angles. The projectile launched at the greater angle will have a velocity vector with a greater vertical component and a smaller horizontal component. • The greater vertical component results in a higher path; the smaller horizontal component results in less range (distance).

Example Question: Herman the human cannonball is launched from level ground at an angle of 30° above the horizontal with an initial velocity of 26 m/s. How far does Herman travel horizontally before reuniting with the ground? • Answer: Our first step in solving this type of problem is to determine Herman's initial horizontal and vertical velocity. We do this by breaking up his initial velocity into vertical and horizontal components: • vi(x) = vi * cos() = 26(m/s) * cos(30 o) = 22. 5 m/s • vi(y) = vi * sin() = 26(m/s) * sin(30 o) = 13 m/s

• Next, we'll analyze Herman's vertical motion to find out how long he is in the air. We'll analyze his motion on the way up, find the time, and double that to find his total time in the air: List Variables • vi=13 m/s • vf=0 • d=? • a=-9. 81 m/s 2 • t=? Solving For t t 1/2= vfy-viy/-a t 1/2= 0 -13. 0 m/s/-9. 8 m/s*s = 1. 3265 sec t= 1. 33*2 = 2. 65 sec

• Now that we know Herman was in the air 2. 65 s, we can find how far he moved horizontally, using his initial horizontal velocity of 22. 5 m/s. • • vi=22. 5 m/s • vf=22. 5 m/s • d=? • a=0 • t=2. 65 s Answer vx=d/t so d(x) = vx * t or d = (22. 5 m/s*2. 65 s)=59. 6 m • Therefore, Herman must have traveled 59. 6 m horizontally before returning to the Earth.

Question For projectile shot at an angle of 60 degrees. What component is being depicted in the following graphs and why?

Answer • Each of these graphs show acceleration. This would have to be the y-axis because there is no acceleration on the x-axis.

Question • You should have found that launch angle is a key variable. – What angle causes the largest height? – What angle causes the largest range?

Answers • 90 degrees to the ground causes the greatest height. • 45 degrees causes the greatest range.

Practice Problems 1. ) A cannon ball is fired at an angle of 36 degrees with an initial velocity of 20 m/s. What are the vertical and horizontal components? 2) A cannon ball is fired at an angle of 36 degrees with an initial velocity of 20 m/s. How long is the ball in the air? 3) A cannon ball is fired at an angle of 36 degrees with an initial velocity of 20 m/s. How far does the ball travel?

Answers 1. ) vi(x) = vi * cos(theta) = 20(m/s) * cos(36 o) = 11. 8 m/s vi(y) = vi * sin(theta) = 20(m/s) * sin(36 o) = 16. 2 m/s 2) t 1/2= 0 -16. 2 m/s/-9. 8 m/s*s = 1. 65 sec * 2 = 3. 3 sec 3) dx= 3. 3 sec * 11. 8 m/s = 39. 0 m

Learning Projectile Motion • • • Learning Objectives Your Learning Goal is to be able to: ● Predict how varying conditions will effect the path of a projectile and explain your reasoning. ● Use the Ph. ET simulation to confirm or disprove your predictions. ● Explain why your predictions were correct or incorrect. Your Grade! - Click here for the worksheet that you will turn in for credit. Projectile_Motion_WS Click Here for steps to solving Projectile Motion Problems. Projectile_Motion_Steps Click Here to see vector notes and homework answers. Vector_Notes_Homework References: http: //aplusphysics. com/courses/regents/kinematics/regents_projectile_motion. h tml http: //phet. colorado. edu/en/simulation/projectile-motion