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What is our frame of reference?,[object Object],Sign Changes,[object Object],Acceleration and velocity at the top,[object Object],X vs t, V vs t, A vs t,[object Object],Complicated Free Fall,[object Object]
We need to determine our origin.,[object Object],Multiple points to choose from,[object Object],Generally pick the ground,[object Object],Sign changes,[object Object],Acceleration is negative,[object Object],Velocity changes signs,[object Object],Frame of reference,[object Object]
A ball is thrown up in the air. At what point is it’s velocity zero? At what point is its acceleration zero?,[object Object],How will a objects starting velocity compare with it’s final velocity? (just before it hits the ground),[object Object],Conceptual Questions,[object Object]
A ball is thrown up into the air with a velocity of 5 m/s. How long is the ball in the air before it hits the ground?,[object Object],Sample Problem,[object Object]
Solution,[object Object],We know the following,[object Object],V0 = 5m/s,[object Object],Vf = -5m/s,[object Object],A = -9.81 m/s2,[object Object],D = ?,[object Object],T = ?,[object Object],Vf = at + V0,[object Object],-5 = (-9.81)(t) + 5,[object Object]
If a ball is thrown into the air, and reaches a height of 91.5m, how fast was it thrown?,[object Object],Group Problem,[object Object]
Solution,[object Object],a = -9.8 m/s2,[object Object],vf = 0 m/s,[object Object],d = 91.5 m,[object Object],vi = ?? ,[object Object],t = ??,[object Object]
THE MONKEY AND THE ZOOKEPER,[object Object],A monkey spends most of its day hanging from a branch of a tree waiting to be fed by the zookeeper. The zookeeper shoots bananas from a banana cannon. ,[object Object],Unfortunately, the monkey drops from the tree the moment that the banana leaves the muzzle of the cannon and the zookeeper is faced with the dilemma of where to aim the banana cannon in order to feed the monkey. ,[object Object],If the monkey lets go of the tree the moment that the banana is fired, where should he aim the banana cannon? ,[object Object],Wilson/Buffa, COLLEGE PHYSICS, Media Portfolio, 5/e, ©2003. Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey.,[object Object]
If the zookeeper aims above the monkey what would be the path of the banana? Would the banana hit the monkey?,[object Object],Let’s find out!,[object Object],What if there was no gravity?,[object Object],Banana thrown ABOVE the monkey,[object Object],http://www.physicsclassroom.com/mmedia/vectors/mzg.html,[object Object],Wrong move!,[object Object]
If the zookeeper aims directly at the monkey what would be the path of the banana? Would the banana hit the monkey?,[object Object],Let’s find out!,[object Object],Banana thrown AT the monkey,[object Object],http://www.physicsclassroom.com/mmedia/vectors/mzf.html,[object Object],What will happen if the banana is thrown at a lower speed?,[object Object],Banana thrown slowly at the monkeyhttp://www.physicsclassroom.com/mmedia/vectors/mzs.html,[object Object],Happy monkey!,[object Object],Can you ever miss the monkey? Why?,[object Object]
PROJECTILE MOTION,[object Object],An object launched into space without motive power of its own is called a projectile. If we neglect air resistance, the only force acting on a projectile is its weight, which causes its path to deviate from a straight line. ,[object Object],The projectile has a constant horizontal velocity and a vertical velocity that changes uniformly under the influence of the acceleration due to gravity.,[object Object]
HORIZONTAL PROJECTION                                                   If an object is projected horizontally, its motion can best be described by considering its horizontal and vertical motion separately. ,[object Object],In the figure we can see that the vertical velocity and position increase with time as those of a free-falling body. Note that the horizontal distance increases linearly with time, indicating a constant horizontal velocity.,[object Object]
3.1 An airplane traveling at 80 m/s at an elevation of 250 m drops a box of supplies to skiers stranded in a snowstorm. ,[object Object],a. At what horizontal distance from the skiers should the supplies be dropped?,[object Object],vo = 80 m/s,[object Object],y = 250 m,[object Object],y = voyt + ½ gt2,[object Object],= 7 s,[object Object],x = voxt,[object Object],  = 80(7),[object Object],  = 560 m,[object Object]
b. Find the magnitude of the velocity of the box as it reaches the ground.,[object Object],vx = vox = 80 m/s ,[object Object],and,[object Object],vy = voy + gt,[object Object],   = 9.8 (7),[object Object],   = 68.6 m/s,[object Object],= 105.4 m/s,[object Object]
3.2 A person standing on a cliff throws a stone with a horizontal velocity of 15.0 m/s and the stone hits the ground 47.0 m from the base of the cliff.  How high is the cliff? ,[object Object],vo = 15.0 m/s,[object Object],x = 47.0 m,[object Object],vy = 0,[object Object],= 3.13 s,[object Object],y = voy t +½ gt2,[object Object],  = 0 + ½ (9.8)(3.13)2,[object Object],  = 48 m,[object Object]
Vector Review,[object Object]
PROJECTILE MOTION AT AN ANGLE,[object Object],The more general case of projectile motion occurs when the projectile is fired at an angle.,[object Object]
Problem Solution Strategy:,[object Object],1.Upward direction is positive. Acceleration due to gravity (g) is downward thus 	g = - 9.8 m/s2,[object Object],2.Resolve the initial velocity vo into its x and y components:,[object Object],vox = vo cos θ	voy = vo sin θ,[object Object],3.The horizontal and vertical components of its positionat any instant is given by:x = voxt	y = voy t +½gt2,[object Object],4.The horizontal and vertical components of itsvelocityat any instant are given by: vx = vox 	vy = voy + gt,[object Object],5.The finalposition and velocity can then be obtained from their components. ,[object Object]
3.3 An artillery shell is fired with an initial velocity of 100 m/s at an angle of 30 above the horizontal. Find:,[object Object],a. Its position and velocity after 8 s,[object Object],vox = 100 cos 30,[object Object],     = 86.6 m/s,[object Object],voy = 100 sin 30,[object Object],     = 50 m/s,[object Object],vo = 100 m/s, 30,[object Object],t = 8 s,[object Object],g = - 9.8 m/s2,[object Object],x = vox t,[object Object],   = 86.6(8),[object Object],   = 692.8 m,[object Object],y = voy t + ½ gt2,[object Object],   = 50(8) + ½ (-9.8)(8)2,[object Object],   = 86.4 m,[object Object],vx = vox = 86.6 m/s,[object Object],vy = voy + gt,[object Object],    = 50 + (-9.8)(8),[object Object],    = - 28.4 m/s,[object Object]
b. The time required to reach its maximum height,[object Object],At the top of the path: ,[object Object],vy = 0,[object Object],vy = voy + gt ,[object Object],= 5.1 s,[object Object],c. The horizontal distance (range) ,[object Object],Total time ,[object Object],T = 2t,[object Object],    = 2(5.1),[object Object],    = 10.2 s,[object Object],x = vox t,[object Object],   = 86.6(10.2),[object Object],   = 883.7 m,[object Object]
3.4 A plastic ball that is released with a velocity of 15 m/s stays in the air for 2.0 s.,[object Object],a. At what angle with respect to the horizontal was it released?,[object Object],vo = 15 m/s,[object Object],t = 2 s,[object Object],time to maximum height = 1 s,[object Object],at the top vy = 0,[object Object],vy = voy + gt,[object Object],=40.8º,[object Object]
b. What was the maximum height achieved by the ball?,[object Object],y = voy t +½gt2,[object Object],   = (15)(sin40.8º)(1) + ½(-9.8)(1)2,[object Object],   = 4.9 m,[object Object]
3.5 An arrow was shot at an angle of 55º with respect to the horizontal. The arrow landed at a horizontal distance of 875 m. Find the velocity of the arrow at the top of its path.,[object Object],θ = 55º,[object Object],x = 875 m,[object Object],total time of flight:,[object Object],y = voy t +½gt2,[object Object],if y = 0,[object Object],0 = t (voy  + ½ gt),[object Object],voy = ½ gt ,[object Object],vo  sin θ  =½ g t,[object Object]
 substituting the time and solving for vo,[object Object],= 95.5 m/s,[object Object],At the top of its path the arrow has vy = o and,[object Object],vx = vox,[object Object],    = 95.5 cos 55º,[object Object],    = 54.8 m/s,[object Object]
3.6 Find the range of a gun which fires a shell with muzzle velocity vo at an angle θ . What is the maximum range possible?,[object Object],At top vy = 0,[object Object],vy = voy + gt,[object Object],    = vo sin θ - gt,[object Object],x = vxt,[object Object],Total time  = 2t ,[object Object]
sin θ cos θ= ½ sin 2θ,[object Object],Maximum range is 45 since 2θ = 90,[object Object]
b. Find the angle of elevation   of a gun that fires a shell with muzzle velocity of 125 m/s and hits a target on the same level but 1.55 km distant.,[object Object],vo = 125 m/s,[object Object],x = 1.55 km,[object Object],= 0.9721,[object Object],sin-1(2θ) =76.4,[object Object],           θ  = 38.2º,[object Object]
Projectile Motion
Projectile Motion

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Projectile Motion

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