Zero Launch Angle

since θ=0, then voy=0 and vox= vo

xo=0, yo =h

independent of vo !!

h

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x

y and based on our coordinate system we have

The time required to reach the water 

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Eliminating t and solving for y as a function of x:

The landing point can be found by setting y = 0 and solving for x:

Combining

x

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You drop a package from a plane flying at constant speed in a straight line.  Without air resistance, the package will:

a) quickly lag behind the plane while falling

b) remain vertically under the plane while falling

c) move ahead of the plane while falling

d) not fall at all

Clicker

a) quickly lag behind the plane while falling

b) remain vertically under the plane while falling

c) move ahead of the plane while falling

d) not fall at all

Both the plane and the package have

the same horizontal velocity at the

moment of release. They will maintain

this velocity in the x-direction, so they

stay aligned.

You drop a package from a 

plane flying at constant speed 

in a straight line.  Without air 

resistance, the package will:

Airplane moving with a constant speed at some altitude. In the course of its flight, the plane drops a package from its luggage compartment. 

Motion Airplane and package before being droppedx= xo + voxt Motion of package

after being droppedx= xo + voxty= yo + voyt -1/2 gt2

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Follow-up: what would happen if air resistance is present?

Recall relationship: v0x = v0 cosθ and v0y = v0 sinθ

This gives the equations of motion for the case x(t=0)=0 , y(t=0) = 0:

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We return to our basic equations for general projectile motion

Snapshots of a trajectory; 

For θ=35o vo = 20m/s

red dots are at times

t = 0.5s, t = 1.0s, and t = 1.5s

0.5sec,

1.0sec,

1.5sec,

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The total velocity at any point is found by vectorially adding the vertical component of the velocity, at that point, to the horizontal component of the velocity at that point.

The horizontal velocity remains constant, because there is no acceleration in that direction.

For a non‐zero launch angle, the initial vertical velocity is upward. However the vertical velocity gets smaller and smaller due to the downward acceleration of gravity 

V Horizontal

V Vertical V total

V Horizontal

V VerticalV total

After reaching the maximum height, the projectile follows a zero launch angle trajectory.  The vertical velocity is downward and the speed gets larger and larger. 

If t=0 during launch, this is time at the highest point

Solve for t 

y component of velocity = 0 at the highest point

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Time is independent of the horizontal  

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Find the total elapse time T the projectile remains in the air

Solving for t we find 

If y=0 at launch time, we find the time when the projectile returns to the ground at y=0

Total time is also independent of the horizontal component 

Ignoring the air

resistance, which

of the three punts

has the longest

time in the air

(hang time) ?

Punts 

d) all three have the same hang time

a) b) c)

h

Clicker

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Range: the horizontal distance a projectile travels

If the initial and final elevation are the same:

y=0, solve for t

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Assume the Projectile starts at the origin at t=0

Time when the projectile land (or returns to y=0)

Double solution function 0 to 90 degrees

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Launch Angle (Deg) Launch Angle (Deg)

Complementary angles will produce the same range.– The maximum height will be different for the two angles.– The times of the flight will be different for the two angles.

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Easy to see when the projectile initial angle is 45 the range is a maximum, however for range values less than the maximum there are 2 possible initial projectile angles and consequently 2 different trajectory paths. 

For example, 30 and 60 degrees projectiles find the same mark. The horizontal velocity component is greater for the 30 degree than the 60, however spends less time in the air. For 60 degrees, the trajectory spends more time in the air than the 30, but the horizontal velocity component is less than the 60.