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PROJECTILE MOTION B2.2
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Projectile Motion

Jan 02, 2016

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Benjamin Barber

Projectile Motion. B2.2. Projectile Motion. Projectiles follow curved (parabolic) paths know as trajectories These paths are the result of two, independent motions Horizontally, the motion has uniform velocity (ignoring air resistance) - PowerPoint PPT Presentation
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Page 1: Projectile Motion

PROJECTILE MOTIONB2.2

Page 2: Projectile Motion

PROJECTILE MOTION Projectiles follow curved (parabolic)

paths know as trajectories These paths are the result of two,

independent motions Horizontally, the motion has uniform

velocity (ignoring air resistance) Vertically, the motion has uniform

acceleration (due to gravity)

Page 3: Projectile Motion

FORMULAE All the formulae for uniform acceleration

may be used to help solve projectile problems

These formula will be applied to horizontal and vertical situations independently

Component vectors will aid us in determining the given values for variables in our formulae.

Page 4: Projectile Motion

CAR OFF CLIFF

A car driving straight off of a cliff will take the same amount of time to hit the ground as an identical car being dropped vertically beside it

The distance the car lands from the base of the cliff is determined by the time gravity takes to bring the car down and the horizontal speed at which the car moves away from the cliff

Page 5: Projectile Motion

RANGE OF A CANNON

The range (firing distance) of a cannon is dependant on the time the cannonball has in the air.

The longer it is in the air, the further it will travel.

Aiming upward (to a point) helps lengthen this “hang time” and increases range

The initial vertical component of velocity and gravity together determine the “hang time”

Page 6: Projectile Motion

RANGE (IGNORING AIR RESISTANCE)

Page 7: Projectile Motion

PROBLEM #1 A rock is thrown horizontally at 10.0m/s

from the top of a cliff 122.5m high.How long does it take the rock to reach the

ground?What is the horizontal displacement of the

rock?

Page 8: Projectile Motion

SOLUTION #1 The method here is to solve the vertical

portion of the question first, in order to find out the “hang time”

If one knows the time the rock has in the air, and we know it covers horizontal ground at 10m/s, we can calculate it’s displacement easily.

Page 9: Projectile Motion

SOLUTION #1

HorizontallyGiven:Uniform MotionV1=10.0 m/s [right]

t=?Dd=?

VerticallyGiven:Uniform AccelerationV1=0 m/s

V2=? (not needed)

A = -9.8m/s2

t=?Dd=122.5m

Page 10: Projectile Motion

SOLUTION #1

This formula is chosen because it excludes v2, since we don’t have it, nor are we asked to find it.

Since v1 is zero, the first term disappears Solve this equation for Dt and use it with

the horizontal data to find Dd.

Page 11: Projectile Motion

PROBLEM #2 Determine an equation for the range of

a projectile (ignoring air resistance) with an initial launch speed of v, launched at an angle of q from the horizontal when the acceleration due to gravity is g.