Projectile MotionBy Allison Appleby

For Physics 2011-12References: Pearson Aust (2010) In2PhysicsShadwick, B (2003) Surfing Physics: Space. Science PressAndriessen et al (2008) Physics 2 HSC course; 3rd edition. John Wiley and Sons Aus Ltd

Projectile Motion Ideas through time

Before Galileo- ideas of Aristotle: horizontal motion and then vertical drop

Definition

Projectile Motion Ideas through time

After Galileo- projectile path is part of a parabola with separate horizontal and vertical components (VECTORS)

Horizontal motion is constant velocity

Vertical motion is accelerating (g)Definition

Projectile Motion Ideas through time

When these components are put together we get parabolic motion

Projectile Motion Ideas through time

The motion of an object depends on the FRAME OF REFERENCE

The motion of the object from a viewer at a distance

A person running with the object would only see the vertical component of motion

Remember from preliminary physics:

The motion of a moving object relative to another moving object:

vB= vB-vA

This is also called the Galilean Transformation

Example

20km/hrClick for answer

Ideal parabolic trajectory

•Air resistance must be negligible

•Height and range of motion is small enough that the curvature of the earth can be ignored

•Vertical component is the y axis- acceleration due to gravity ay=g• ↑ is the positive direction• ↓ is the negative direction• As gravity is down ay= -9.8 m/s2

•Horizontal component is the x axis- velocity is constant ax=0• → is the positive direction• ← is the negative direction

Properties of ideal parabolic trajectories

•At maximum height vy=0•Trajectory is horizontally symmetrical about the maximum height•It takes the same time to reach maximum height as it does to fall back to the original height•Initial speed= final speed (on horizontal ground)•Maximum height is reached at 90o launch angle and maximum range is reached at 45o launch angle•All objects projected horizontally from the same height have the same time of flight as one dropped from rest at the same height (initial vertical velocity = 0)

Solving Projectile Motion ProblemsWe use SUVAT equations from preliminary physics:

s = r = displacement = Δx = xf = xi = Δy = yf – yi

SOHCAHTOAv=s/t or vav= Δ r / Δ t

aav= v-u t

SUVAT (straight line motion)

Horizontal component

Vertical component

ux=ucosΘ uy=usinΘ

v = u + at vx=ux (ax=0) vy=uy + ayt

v2=u2 + 2as vx2=ux 2 vy

2=uy 2 + 2ayΔy

s= ut + ½ at2 Δx = ux t Δy = uy t + ½ ayt