2Projectile Examples Hockey puck Tennis ball Basketball Golf ball VolleyballArrowShot putJavelinTennis ballGolf ballFootballSoftballSoccer ballBulletThese are all examples of things that areprojected, then go off under theinfluence of gravity
3Not projectiles Jet plane Rocket Car (unless it looses contact with ground)
4Understanding Projectiles The key to understandingprojectile motion is to realizethat gravity acts vertically it affects only the verticalpart of the motion, not thehorizontal part of the motion
5DemonstrationWe can see that the horizontal and vertical motions are independentThe red ball falls verticallyThe yellow ball was given a kick to the right.They track each other vertically step for step and hit the ground at the same time
6Projectile Paths In the absence of gravity a bullet would follow a straight line forever.With gravity it FALLS AWAY fromthat straight line!
8Sample ProblemA zookeeper finds an escaped monkey hanging from a light pole. Aiming her tranquilizer gun at the monkey, she kneels 10.0 m from the light pole,which is 5.00 m high. The tip of her gun is 1.00 m above the ground. At the same moment that the monkey drops a banana, the zookeeper shoots. If the dart travels at 50.0 m/s,will the dart hit the monkey, the banana, or neither one?
9Sample Problem1 . Select a coordinate system. The positive y-axis points up, and the positive x-axis points along the ground toward the pole. Because the dart leaves the gun at a height of 1.00 m, the vertical distance is 4.00 m.
10Sample Problem2 . Use the inverse tangent function to find the angle that the initial velocity makes with the x-axis.
113 . Choose a kinematic equation to solve for time. Sample Problem3 . Choose a kinematic equation to solve for time.Rearrange the equation for motion along the x-axis to isolate the unknown Dt, which is the time the dart takes to travel the horizontal distance.
12Sample Problem4 . Find out how far each object will fall during this time. Use the free-fall kinematic equation in both cases.For the banana, vi = 0. Thus:Dyb = ½ay(Dt)2 = ½(–9.81 m/s2)(0.215 s)2 = –0.227 m
13Sample ProblemThe dart has an initial vertical component of velocity equal to vi sin q, so:Dyd = (vi sin q)(Dt) + ½ay(Dt)2Dyd = (50.0 m/s)(sin 21.8)(0.215 s) +½(–9.81 m/s2)(0.215 s)2Dyd = 3.99 m – m = 3.76 m
14Sample Problem 5 . Analyze the results. Find the final height of both the banana and the dart.ybanana, f = yb,i+ Dyb = 5.00 m + (–0.227 m)ybanana, f = 4.77 m above the ground
15Sample Problem ydart, f = yd,i+ Dyd = 1.00 m + 3.76 m ydart, f = 4.76 m above the groundThe dart hits the banana.The slight difference is due to rounding.
17Newton’s First Law of Motion “Every object continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state of motion by forces impressed upon it ”The tendency of matter to maintain its state of motion is known as INERTIA.
23Projectile motion – key points The projectile has both a vertical and horizontal component of velocityThe only force acting on the projectile once it is shot is gravity (neglecting air resistance)At all times the acceleration of the projectile is g = 9.8 m/s2 downwardThe horizontal velocity of the projectile does not change throughout the path
24Key points, continuedOn the rising portion of the path gravity causes the vertical component of velocity to get smaller and smallerAt the very top of the path the vertical component of velocity is ZEROOn the falling portion of the path the vertical velocity increases
26More key pointsIf the projectile lands at the same elevation as its starting point it will have the same vertical SPEED as it began withThe time it takes to get to the top of its path is the same as the time to get from the top back to the ground.The range of the projectile (where it lands) depends on its initial speed and angle of elevation
28Sample ProblemA 2.00 m tall basketball player wants to make a basket froma distance of 10.0 m. If he shoots the ball at a 450 angle, atwhat initial speed must he throw the ball so that it goesthrough the hoop without striking the backboard?yxy0
30Maximum RangeWhen an artillery shell is fired the initial speed of the projectile depends on the explosive charge – this cannot be changedThe only control you have is over the angle of elevation.You can control the range (where it lands) by changing the angle of elevationTo get maximum range set the angle to 45°