PROJECTILE MOTION

Projectile Examples Hockey puck Tennis ball Basketball Golf ball Volleyball Arrow Shot put Javelin Tennis ball Golf ball Football Softball Soccer ball Bullet These are all examples of things that are projected, then go off under the influence of gravity

Not projectiles Jet plane Rocket Car (unless it looses contact with ground)

 The key to understanding projectile motion is to realize that gravity acts vertically  it affects only the vertical part of the motion, not the horizontal part of the motion

Demonstration We can see that the horizontal and vertical motions are independent The red ball falls vertically The 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

In the absence of gravity a bullet would follow a straight line forever. With gravity it FALLS AWAY from that straight line!

Shoot the Monkey

Sample Problem A 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?

1 . 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.

2 . Use the inverse tangent function to find the angle that the initial velocity makes with the x-axis.

3 . 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.

4. Find out how far each object will fall during this time 4 . 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 The dart has an initial vertical component of velocity equal to vi sin q, so: Dyd = (vi sin q)(Dt) + ½ay(Dt)2 Dyd = (50.0 m/s)(sin 21.8)(0.215 s) +½(–9.81 m/s2)(0.215 s)2 Dyd = 3.99 m – 0.227 m = 3.76 m

5 . Analyze the results. ydart, f = yd,i+ Dyd = 1.00 m + 3.76 m 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 ydart, f = yd,i+ Dyd = 1.00 m + 3.76 m ydart, f = 4.76 m above the ground The dart hits the banana. The slight difference is due to rounding.

Football without gravity

No gravity is good for kickers

Basketball – without gravity

Hitting the target – aim high, not directly at the target BULLSEYE!

Path of the Projectile g falling rising Height v Distance downfield (range) Horizontal velocity Vertical velocity v projectile

Horizontal Motion

Vertical Motion

Projectile motion – key points The projectile has both a vertical and horizontal component of velocity The 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 downward The horizontal velocity of the projectile does not change throughout the path

Key points, continued On the rising portion of the path gravity causes the vertical component of velocity to get smaller and smaller At the very top of the path the vertical component of velocity is ZERO On the falling portion of the path the vertical velocity increases

More key points If the projectile lands at the same elevation as its starting point it will have the same vertical SPEED as it began with The 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

Example: A 2.00 m tall basketball player wants to make a basket from a distance of 10.0 m. If he shoots the ball at a 450 angle, at what initial speed must he throw the ball so that it goes through the hoop without striking the backboard? y x y0

Equations to Choose from

Maximum Range When an artillery shell is fired the initial speed of the projectile depends on the explosive charge – this cannot be changed The only control you have is over the angle of elevation. You can control the range (where it lands) by changing the angle of elevation To get maximum range set the angle to 45°

Interactive http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html http://jersey.uoregon.edu/vlab/Cannon/

The ultimate projectile: Putting an object into orbit Imagine trying to throw a rock around the world. If you give it a large horizontal velocity, it will go into orbit around the earth!

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