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Projectile Motion Horizontally
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Projectile Motion Once a difficult problem for cannoneers
If the vertical and horizontal components are separated, the problem becomes simple.
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Projectile Motion Horizontal – simple as a ball rolling across the table Use equations like x = v x t
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Projectile Motion Vertical – simple as a free falling gravity problem
Use equations ∆y = (1/2)gt2 and vf = gt
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Projectile Motion NOTE: Air and other resistance is being ignored for the time being.
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Projectile Motion
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Projectile Motion of a Cannon Ball Projected Horizontally
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Question 1 Consider these diagrams to answer the following questions:
a. Which shows the initial velocity? a b. Which shows the acceleration vector? b
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Question 2 What determines the time for the cannon ball to hit the ground?
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Question 2 Solution The time for the cannon ball to hit the ground is determined by the height from which it drops.
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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
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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
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Simulation: Constant Velocity in the Horizontal Direction
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Horizontal & Vertical Velocities
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Horizontal & Vertical Displacements
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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 m/s2 Velocity (Constant or Changing) Constant Changing (by m/s each second)
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Projectile Motion at Zero Angle
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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?
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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
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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?
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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
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Characteristics & Assumptions
Planar motion Air resistance: negligible Gravity: downward No horizontal acceleration Constant vertical acceleration Parabolic trajectory Independent horizontal and vertical motions
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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?
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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
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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?
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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
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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.
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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
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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.
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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
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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?
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Question 9 A toy car runs of the edge of a table that is m high. If the car lands 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?
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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?
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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?
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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?
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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?
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Monkey & Hunter Experiment
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PROJECTILE MOTION
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Projectile Motion
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Projectile Motion Velocity
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Projectile Motion
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Projectile Motion
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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
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Projectile Motion Air Resistance
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Monkey and the Hunter
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Analysis of Projectile Motion
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Graphs for a Projectile Motion
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Range R Simulation
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Terminal Velocity Sky-diver in free fall
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Terminal Velocity Abc Terminal velocity is reached when f(v) = g
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Terminal Velocity: v-t Graph
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