a projectile is an object that has been given an initial velocity by some sort of short-lived force,...

A projectile is an object that has been given an initial velocity by some sort of short-lived force, and then moves through the air under the influence of gravity.Baseballs, stones, or bullets are all examples of projectiles.

Topic 2.1 ExtendedI – Projectile motion

You know that all objects moving through air feel an air resistance (recall sticking your hand out of the window of a moving car).We call that air resistance drag and will IGNORE it for the time being. The drag force is a complicating factor that we are not yet ready to deal with.

FYI: In the analysis that follows, we will NOT consider the drag force. Keep in the back of your mind that what we derive will be precise only in the absence of air. In air, our results will only be a good, first-order APPROXIMATION.

It is important to note that the acceleration of a projectile is caused by the gravitational attraction of the earth to the object.

Thus, while there IS an acceleration in the y-direc-tion, there NOT an acceleration in the x-direction.


Pro

ject

ile s

low

ing

dow

n in

+y

dir. P

rojectile speeding up in -y dir.

Projectile moving at constant speed in +x dir. ax = 0ay = -g ay = -g

The kinematic equations are used for projectile motion, because the accelerations are constant.The acceleration in the x-direction is 0, and the acceleration in the y-direction is -g. Thus


vx = v0x + axtx = x0 + v0xt + axt21

2

vx2 = v0x

2 + 2axx

x = x0 + v0xt

vx = v0x

vy = v0y + ayty = y0 + v0yt + ayt21

2

vy2 = v0y

2 + 2ayy

y = y0 + v0yt - gt212

vy = v0y - gt

vy2 = v0y

2 + 2gy

General kinematic eq'ns in x

General kinematic eq'ns in y

Projectile eq'ns in x

Projectile eq'ns in y

Projectile Motion

Going back to the original baseball, we find the values of v0x and v0y from the velocity vector v0:Recall that v is ALWAYS tangent to the path of the projectile. Thus v0 is tangent to the path at t0:


v0

0

v0x = v0 cos 0

v 0y

= v

0 si

n 0

We use the velocity triangle to resolve the components of v0: Now our projectile motion equations are complete.

v0x = v0 cos 0

v0y = v0 sin 0

Projectile Motion

Consider a cannonball that is given an initial velocity of 50.0 m/s at an elevation angle of 30.0°.


30°x

y

50 m/s

v0

(a) What are v0x and v0y?

= 50.0 sin 30.0°= 25.0 m/s

= 50.0 cos 30.0°= 43.3 m/s

v0y = v0 sin 0

v0x = v0 cos 0



(b) What are the tailored equations of motion of the cannonball?

x = x0 + v0xt

vx = v0x

x = 43.3t

vx = 43.3

y = y0 + v0yt - gt212

vy = v0y - gt

vy2 = v0y

2 + 2gy

y = 25.0t - 4.90t2

vy = 25.0 - 9.80t

vy2 = 25.02 - 19.6y

Tailored Equations



(c) What maximum height above the muzzle will the cannonball reach?

The height is determined using the y equations:

vy = 25.0 - 9.80t0 = 25.0 - 9.80t

9.80t = 25.0t = 2.55 s

Now substitute:

y = 25.0t - 4.90t2

y = 25.0(2.55) - 4.9(2.55)2

y = 31.9 m

Method 1:

Question: How do we know vy = 0 at the maximum height?

Method 2:vy

2 = 25.02 - 19.6y

02 = 25.02 - 19.6y

19.6y = 25.02

y = 31.9 m

FYI: If all you need is y, Method 2 is the best.



(d) How long does it take the cannonball to come back down to its original muzzle height?

The time is determined using the y equations:

y = 25.0t - 4.90t2

0 = 25.0t - 4.90t2

Question: How do we know y = 0 at the desired time?

4.90t2 = 25.0t

4.90t = 25.0

t = 5.10 s

Method 1: Method 2:vy

2 = 25.02 - 19.6yvy

2 = 25.02 - 19.6(0)

vy2 = 25.02

vy = ±25.0vy = -25.0

Question: How do we know vy is NEGATIVE at the desired time?

vy = 25.0 - 9.80t-25.0 = 25.0 - 9.80t9.80t = 50.0

t = 5.10 s



(e) How far does the cannonball travel before reaching its original muzzle height?

Distance is determined using the x equations:We know it takes to 5.10 s to reach this distance.

x = 43.3t

x = 43.3(5.10)

x = 221 m



(f) In what direction is the cannonball traveling when it reaches its original muzzle height?

Direction is determined using the velocity equations:

We know it takes to 5.10 s to reach this position.

vx = 43.3 m/s

vy = 25.0 - 9.80t

vy = 25.0 - 9.80(5.10)

vy = -25.0 m/s

vx = 43.3 m/s

vy = -25.0 m/s

v

= tan-1vy

vx

= tan-1 25.043.3

= 30.0°



(g) What is the displacement of the cannonball at t = 2.00 seconds?

Displacement is determined using the position equations:

x = 43.3t

y = 25.0t - 4.90t2

x = 43.3(2.00)

x = 86.6 m

y = 25.0(2.00) - 4.90(2.00)2

y = 30.4 m

x = x - x0 = 86.6 m

y = y - y0 = 30.4 m

THE RANGE EQUATION


Sometimes a projectile begins and ends its flight at the same height. If such is the case we can derive a simple formula that tells us the range R of the projectile as a function of initial velocity v0 and elevation angle (or projection angle) 0 .

x

y

v0

R

θ0

From the projectile equations we get...

x = x0 + v0xty = y0 + v0yt - gt212

R = 0 + v0(cos θ0)t0 = 0 + v0(sin θ0)t - gt212

R = v0(cos θ0)tgt2 = 2v0(sin θ0)t

t = 2v0(sinθ0)g R = (2 sinθ0 cosθ0)

v02 g

R = sin 2θ0 v0

2 g

the range equation

R = v0(cos θ0)2v0(sinθ0)

g

FYI: Recall the trigonometric identity 2 sin cos sin 2

FYI: sin 2 has a maximum value of 1. Therefore R has a maximum value of Rmax = v0

2/g when 0 = 45°.

This equation is only valid if beginning and ending heights are the same.

THE RANGE EQUATION


What angle of elevation (projection angle) should a port cannon have in order to hit a pirate ship 800. m off shore. Assume the muzzle velocity of the cannon ball is 150. m/s.

the “level”

R = 800 m

θ0

R = sin2θ0 v0

2 g

800 = sin 2θ0 1502 10

v0 = 150 m/s

FYI: We need to find 0.

From the range equation we get

THE RANGE EQUATION



800 = sin 2θ0 1502 10

= sin 2θ0 800(10)

1502

sin 2θ0 = 0.3556 x

y

2θ0 = 20.83° 2θ0 = 180° - 20.83°

2θ0 = 159.17°

θ0 = 79.59°

θ0 = 10.42°

2θ0 2θ0

This angle also has the same

sine of 0.3556.

One possibility Another possibility

opp opphyphyp

SOH

THE RANGE EQUATION



the “level”

R = 800 m

v0

v0

θ0 =

79.59°

θ0 = 10.42°

Which way takes longest?There are TWO ways to hit the ship.

(a) How long is the stone in the air?


Suppose you throw a stone at a 30.0° angle and a speed of 50.0 m/s from the top of a bridge that is 40.0 m above a river.

x

y50 m/s

FYI: Since initial and final heights are NOT the same, we CANNOT use the range equation. We will work this problem out from the basic projectile motion equations.

30°

xmax

FYI: Put in a reasonable coordinate system. In this case, if the y-axis begins at the lowest point the ball will ever reach in the problem, you will never have to worry about negative y-values.




First analyze your velocity triangle:

30°x

y

50 m/s

v0

= 50.0 sin 30.0°= 25.0 m/s

= 50.0 cos 30.0°= 43.3 m/s

v0y = v0 sin 0

v0x = v0 cos 0




The flight time is limited by the y-equations:

y = y0 + v0yt - gt212

0 = 40.0 + 25.0t - 4.90t2

-4.90t2 + 25.0t + 40.0 = 0 a b ct =

-b ± b2 - 4ac2a

=-25 + 37.54

-9.8

=-25 ± 252 - 4(-4.9)(40)

2(-4.9)

=-25 ± 37.54

-9.8=

-25 - 37.54-9.8

= -1.28 s

= 6.38 sMethod 1:




Some people prefer to NOT use the quadratic formula: Method 2:

vy2 = v0y

2 + 2gy vy2 = 25.02 - 19.6(0-40)

vy2 = 1409

vy = -37.54

FYI: Why MUST you choose the NEGATIVE vy?

vy = v0y - gt vy = 25.0 - 9.80t-37.54 = 25.0 - 9.80t

9.80t = 62.54

t = 6.38 s

(b) How far along the river does the stone go?



For horizontal distance use the equations in x:

x = x0 + v0xt x = 43.3t

x = 43.3(6.38)

x = 276 m


Suppose an airplane traveling at 144 km/h at an altitude of 400 m is to make an emergency drop of a package to a group stranded on an island. How far down range must the package be dropped in order to land on the island?

x

y0 = 400 m

144 =kmh

144 kmh

1000 m1 km

1 h3600 s

= 40 m/s

FYI: The original velocity of the projectile is HORIZONTAL, and the same speed as the plane.

FYI: Convert the speed of the plane into compatible units (m/s).

v0 = 40 m/s

v0x = 40 m/s v0y = 0 m/s

x0 = 0 m y0 = 400 m

ax = 0 m/s2 ay = -9.8 m/s2

The initial values


Suppose an airplane traveling at 144 km/h at an altitude of 400 m is to make an emergency drop of a package to a group stranded on an island. How far down range must the package be dropped in order to land on the island?

x

y0 = 400 m

v0 = 40 m/s

y = 400 – 5t2

0 = 400 – 5t2

5t2 = 400

t2 = 80

t = 8.94 s

x = 40tx = 40(8.94)

= 358 m

tim

e li

mit

ran

ge

FYI: The package must be dropped 358 m down-range of the target.

a projectile is an object that has been given an initial velocity by some sort of short-lived force,...

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