Projectile Motion Horizontally Projectile Motion Once a difficult

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)gt 2 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 No (Present? - Yes or No) If present, what direction? ) Velocity (Constant or Changing) Yes “g” is downward at -9. 81 m/s 2 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 � Height of table � ∆y = vit + ½ gt 2 � ∆y = ½ (-9. 81 m/s 2)(4. 0 s)2 � ∆y = -78 m……… 78 m

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 � 30 speed is constant. m/s = vx �Vertical Speed �vf = vi + gt �vf = (-9. 81 m/s 2 )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 at what height will the ball drop after each second of motion? �

Solution �∆y = vit + ½ gt 2 �-80. m = ½ (-9. 81 m/s 2)t 2 �-160 m = (-9. 81 m/s 2) t 2 �t = 4. 0 s �Each sec the displacements are as follows: ∆y = vit + ½ gt 2 = ½ (-9. 81 m/s 2)(1 sec)2 �-4. 9 m, -20 m, -44 m, -78 m

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 at what height will the ball be after each second of travel? �

Question 5 Solution �∆y = vit + ½ gt 2 �-80. m = ½ (-9. 81 m/s 2)t 2 �-160 m = (-9. 81 m/s 2) t 2 �t = 4. 0 s �Each sec the displacements are as follows: ∆y = vit + ½ gt 2 = ½ (-9. 81 m/s 2)(1 sec)2 �-4. 9 m, -20 m, -44 m, -78 m

�A Question 6 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 gt 2 �-. 60 m = ½ (-9. 81 m/s 2)t 2 �-1. 20 m = (-9. 81 m/s 2) t 2 �t =. 35 s �Horizontal Distance �x = v (t) = 2. 4 m/s (. 3497 s) =. 84 m

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 gt 2 �-22 m = ½ (-9. 81 m/s 2)t 2 �-44 m = (-9. 81 m/s 2) t 2 �t = 2. 1 s �Initial Horizontal Velocity �vx = x/t = 35 m/2. 1178 s = 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