Projectile Motion Horizontally

Projectile Motion Once a difficult problem for cannoneers If the vertical and horizontal components are separated, the problem becomes simple.

Projectile Motion Horizontal – simple as a ball rolling across the table Use equations like x = v x t

Projectile Motion Vertical – simple as a free falling gravity problem Use equations ∆y = vi t + (1/2)gt2 and vf = vi + gt

Projectile Motion NOTE: Air and other resistance is being ignored for the time being.

Projectile Motion

Projectile Motion of a Cannon Ball Projected Horizontally

Question 1 Consider these diagrams to answer the following questions: a. Which shows the initial velocity? a b. Which shows the acceleration vector? b

Question 2 What determines the time for the cannon ball to hit the ground?

Question 2 Solution The time for the cannon ball to hit the ground is determined by the height from which it drops.

Question 3 Supposing a snowmobile is equipped with a flare launcher which is capable of launching a sphere vertically (relative to the snowmobile). If the snowmobile is in motion and launches the flare and maintains a constant horizontal velocity after the launch, then where will the flare land (neglect air resistance)? a. in front of the snowmobile b. behind the snowmobile c. in the snowmobile

Solution for Question 3 C The flare will land inside the snowmobile. The flare has the same horizontal velocity as the snowmobile that it was initially in. The velocity does not change

Simulation: Constant Velocity in the Horizontal Direction http://www.physicsclassroom.com/mmedia/vectors/pap.html

Horizontal & Vertical Velocities

Horizontal & Vertical Displacements

Summary of Projectile Motion Horizontally Horizontal Motion Vertical Motion Forces (Present?- Yes or No) If present, what direction No Yes The force of gravity acts downward Acceleration (Present?- Yes or No) If present, what direction?) “g” is downward at -9.81 m/s2 Velocity (Constant or Changing) Constant Changing (by -9.81 m/s each second)

An ugly giant rolls a billiard ball with uniform velocity 30 m/s across the top of his desk. The ball rolls off the edge of the desk and lands on the floor 120 m from the edge of the desk. How high is the table?

Height of Table Time of fall vx = x/t 30 m/s = 120 m / t t = 4.0 s ∆y = vit + ½ gt2 ∆y = ½ (-9.81 m/s2)(4.0s)2 ∆y = -78 m………78m

At what speed will it hit the ground? An ugly giant rolls a billiard ball with uniform velocity 30 m/s across the top of his desk. The ball rolls off the edge of the desk and lands on the floor 120 m from the edge of the desk. At what speed will it hit the ground?

Velocity When It Hits The Ground Horizontally speed is constant. 30 m/s = vx Vertical Speed vf = vi + gt vf = (-9.81 m/s2 )4.0 s vf = 39 m/s Resultant speed = 49 m/s

Characteristics & Assumptions Planar motion Air resistance: negligible Gravity: downward No horizontal acceleration Constant vertical acceleration Parabolic trajectory Independent horizontal and vertical motions

Question 4 Anna Litical drops a ball from rest from the top of 80.-meter high cliff. How much time will it take for the ball to reach the ground and at what height will the ball drop after each second of motion?

Solution ∆y = vit + ½ gt2 -80.m = ½ (-9.81 m/s2)t2 t = 4.0 s Each sec the displacements are as follows: ∆y = vit + ½ gt2 = ½ (-9.81m/s2)(1sec)2 -4.9m, -20m,-44m,-78m

Question 5 A cannonball is launched horizontally from the top of an 80.-meter high cliff. How much time will it take for the ball to reach the ground and at what height will the ball be after each second of travel?

Question 5 Solution ∆y = vit + ½ gt2 -80.m = ½ (-9.81 m/s2)t2 t = 4.0 s Each sec the displacements are as follows: ∆y = vit + ½ gt2 = ½ (-9.81m/s2)(1sec)2 -4.9m, -20m,-44m,-78m

Question 6 A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.

Solution for Question 6 Vertical Time ∆y = vit + ½ gt2 -.60m = ½ (-9.81 m/s2)t2 -1.20m = (-9.81 m/s2) t2 t = .35 s Horizontal Distance x = v (t) = 2.4 m/s ( .3497s) = .84m

Question 7 A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

Solution Question 7 Vertical Time ∆y = vit + ½ gt2 -22m = ½ (-9.81 m/s2)t2 -44m = (-9.81 m/s2) t2 t = 2.1 s Initial Horizontal Velocity vx = x/t = 35 m/2.1178s = 17 m/s

Question 8 You accidentally throw your car keys horizontally at 8.0 m/s from a cliff 64 m high. How far from the base of the cliff should you look for the keys?

Question 9 A toy car runs of the edge of a table that is 1.225 m high. If the car lands 0.400 m from the base of the table, a. how long did it take the car to fall? b. how fast was the car going on the table?

Question 10 Divers in Acapulco dive from a cliff that is 61 m high. If the rocks below the cliff extend outward for 23 m, what is the minimum horizontal velocity a diver must have to clear the rocks?

Question 11 A dart player throws a dart horizontally at a speed of 12.4 m/s. The dart hits the board 0.32 m below the height from which it was thrown. How far away is the player from the board?

Question 12 A ball thrown horizontally from a 13 m high building strikes the ground 5.0 m from the building. With what velocity was the ball thrown?

Question 13 A person leaps horizontally from the top of a tower and lands 17.0 m from the base of the tower. If the speed at which the person was projected was 9.50 m/s, how high is the tower?

Problem Solving Approach Carefully read the problem. List known and unknown information For convenience sake, make a table with horizontal information on one side and vertical information on the other side. Identify the unknown quantity which the problem requests you to solve for. Select either a horizontal or vertical equation to solve for the time of flight of the projectile. With the time determined, use one of the other equations to solve for the unknown. (Usually, if a horizontal equation is used to solve for time, then a vertical equation can be used to solve for the final unknown quantity.)

Projectile Motion

Projectile Motion Velocity

Projectile Motion

Projectile Motion Trajectory Up to a 45 degree angle – greater distance at greater angle After 45 degree – greater height, less distance 45 degree gives the greatest distance Any two angles that add to 90 degrees will hit the same place

Graphs for a Projectile Motion

Terminal Velocity Sky-diver in free fall

Terminal Velocity Abc Terminal velocity is reached when f(v) = g

Terminal Velocity: v-t Graph