more projectile motion discussion: examples. i hope this doesn’t apply to you!

More Projectile Motion Discussion: Examples

More Projectile Motion Discussion: Examples

I hope this doesn’t apply to you!

Solving Projectile Motion Problems1. Read the problem carefully, & choose the object(s) you

are going to analyze.

2. Sketch a diagram.

3. Choose an origin & a coordinate system.

4. Decide on the time interval; this is the same in both directions, & includes only the time the object is moving with constant acceleration g.

5. Solve for the x and y motions separately.

6. List known & unknown quantities. Remember that vx never changes, & that vy = 0 at the highest point.

7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

Example 4.4: Non-Symmetric Projectile Motion

Kinematic Equationsvxi = vicosθi, vyi = visinθi

vxf = vxi , xf = vxi tvyf = vyi - gt

yf = vyi t - (½)gt2

(vyf) 2 = (vyi)2 - 2gyf

A stone is thrown! xi = yi = 0yf = -45.0 m, vi = 20 m/s, θi = 30º

a) Time to hit the ground?b) Speed just before it hits? c) Distance from the base of the

building where it lands?

Example 4.4: SolutionA stone is thrown! xi = yi = 0

yf = -45.0 m, vi = 20 m/s, θi = 30ºa) Time to hit the ground?b) Speed just before it hits? c) Distance from the base of the

building where it lands? First, calculate

vxi = vi cos(θi) = 17.3 m/svyi = vi sin(θi) = 10.0 m/s

a) Time to hit the ground? (Time when yf = -45.0 m)

yf = -45m = vyi t - (½)gt2

A general quadratic must be solved using the quadratic equation! This gives:

t = 4.22 s




(vyf) 2 = (vyi)2 - 2gyf


yf = -45.0 m, vi = 20 m/s, θi = 30ºa) Time to hit the ground?b) Speed just before it hits? c) Distance from the base of the building

where it lands? First, calculate


thit = 4.22 sb) Velocity just before it hits?

vxf = vxi , vyf = vyi – gt so vxf = 17.3 m/svyf = 10 – (9.8)(4.22) = - 31.3 m/s

Speed (vf)2 = (vxf)2 + (vyf)2

vf = 35.8 m/sAngle: tan(θf) = (vyf/vxf) = -(31.3/17.3) = -1.8

θf = -60.9º




(vyf) 2 = (vyi)2 - 2gyf


yf = -45.0 m, vi = 20 m/s, θi = 30ºa) Time to hit the ground?b) Speed just before it hits? c) Distance from the base of the

building where it lands? First, calculate


thit = 4.22 svf = 35.8 m/s, θf = -60.9º

c) Distance from the base of th building where it lands?

xf = vxi thit = (17.3)(4.22) = 73.0 m




(vyf) 2 = (vyi)2 - 2gyf

Example 4.2: The Long JumpA long-jumper leaves the ground at an angle θi = 20° above the horizontal at a speed of vi = 8.0 m/s. a) How far does he jump in the horizontal direction? (Assume his motion is equivalent to that of a particle.) b) What is the maximum height reached?

Kinematic Equationsvxi = vicosθi, vyi = visinθi ,vxf = vxi

xf = vxi t, vyf = vyi – gtyf = vyi t - (½)gt2

(vyf) 2 = (vyi)2 - 2gyf

The Long Jump: SolutionsA long-jumper leaves the ground at an angle θi = 20° above the horizontal at a speed of vi = 8.0 m/s. a) How far does he jump in the horizontal direction? (Assume his motion is equivalent to that of a particle.) b) What is the maximum height reached?



(vyf) 2 = (vyi)2 - 2gyf


a) How far does he jump in the horizontal direction? Range =R = (2vxivyi/g) = 2(7.5)(4)/(9.8)

R = 7.94 m

A long-jumper leaves the ground at an angle θi = 20° above the horizontal at a speed of vi = 8.0 m/s. a) How far does he jump in the horizontal direction? (Assume his motion is equivalent to that of a particle.) b) What is the maximum height reached?



(vyf) 2 = (vyi)2 - 2gyf


R = 7.94 m b) What is the maximum height?

h = [(vyi)2/(2g)]

h = 0.72 m

The Long Jump: Solutions

Example: Driving Off a Cliff!!

vxf = vxi = ? vyf = -gtxf = vxft, yf = - (½)gt2

Time to Bottom: t = √2y/(-g) = 3.19 svx0 = (x/t) = 28.2 m/s

A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are?

Kinematic Equations: vxi = vicosθi, vyi = visinθi ,vxf = vxi xf = vxi t vyf = vyi – gt, yf = vyi t - (½)gt2, (vyf) 2 = (vyi)2 - 2gyf

Solutions: Driving Off a Cliff!!

vxf = vxi = ? vyf = -gtxf = vxft, yf = - (½)gt2

Time to Bottom: t = √2y/(-g) = 3.19 svx0 = (x/t) = 28.2 m/s

A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are?

Kinematic Equations: vxi = vicosθi, vyi = visinθi ,vxf = vxi xf = vxi t vyf = vyi – gt, yf = vyi t - (½)gt2, (vyf) 2 = (vyi)2 - 2gyf

vx = vxi = ?, vyf = -gtxf = vxit, yf = - (½)gt2

Time to the bottom = time when y = - 50 m

- (½)gt2 = - 50 mt = 3.19 s

At that time xf = 90.0 mSo vxi = (xf/t) = (90/3.19)

vxi = 28.2 m/s

Example: Kicked Football

• A football is kicked at an angle θ0 = 37.0° with a velocity of

20.0 m/s, as shown. Calculate:a. The maximum height. b. The time when it hits the ground. c. The total distance traveled in the x direction. d. The velocity at the top. e. The acceleration at the top.

θ0 = 37º, v0 = 20 m/s vx0= v0cos(θ0) = 16 m/s, vy0= v0sin(θ0) = 12 m/s

Conceptual Example

Demonstration!!

vx0

Conceptual Example: Wrong Strategy

“Shooting the Monkey”!!Video Clips!!

Example: A Punt!

vi = 20 m/s, θi = 37º

vxi = vicos(θi) = 16 m/s, vyi= visin(θi) = 12 m/s

Proof that the projectile path is a parabola

xf = vxi t , yf = vyi t – (½)g t2

Note: The same time t enters both equations!

Eliminate t to get y as a function of x.

Solve the x equation for t: t = xf/vxi

Get: yf = vyi (xf/vxi) – (½)g (xf/vxi)2

Or: yf = (vyi /vxi)xf - [(½)g/(vxi)2](xf)2

This is of the form yf = Axf – B(xf)2

A parabola in the x-y plane!!

Example : The Stranded ExplorersProblem: An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers, as shown in the picture. If the plane is traveling horizontally at vi = 42.0 m/s at a height h = 106 m above the ground, where does the package strike the ground relative to the point at which it is released?

vi = 42 m/s

h

Problem

Problem SolutionChoose the origin at ground level, under where the projectile is launched, & up to be the positive y direction. For the projectile:

a. The time to reach the ground is found from the free fall equation, with final height = 0. Choose positive time since the projectile was launched at t = 0.

b. The horizontal range is found from the horizontal motion at constant velocity.

0 65.0 m s ,v 0 35.0 , ,ya g 0 115m,y 0 0 0sin .yv v

2 21 10 0 0 02 2

2 2 120 0 0 0

12

0

0

0 sin

sin sin 49.964s , 2.3655s 9.96s

2

y yy y v t a t y v t gt

v v gt

g

y

0 0cos 65.0 m s cos35.0 9.964s 531mxx v t v t

0 0cos 65.0 m s cos35.0 53.2 m s .xv v

c. At the instant just before the particle reaches the ground, the horizontal component of its velocity is the constant

The vertical component of velocity is found from:

d. The magnitude of the velocity is found from the x and y components calculated in part c. above.

2

0 0 0sin 65.0 m s sin 35.0 9.80 m s 9.964s

60.4 m s

y yv v at v gt

0 0cos 65.0 m s cos35.0 53.2 m s .xv v

2 22 2 53.2 m s 60.4 m s 80.5m sx yv v v

e. The direction of the velocity is

so the object is moving

f. The maximum height above the cliff top reached by the projectile will occur when the y-velocity is 0, and is found from:

1 1 60.4tan tan 48.6

53.2y

x

v

v

48.6 below the horizon .

2 2 2 2

0 0 0 0 max

2 22 2

0 0max 2

2 0 sin 2

65.0 m s sin 35.0sin70.9 m

2 2 9.80 m s

y y yv v a y y v gy

vy

g

more projectile motion discussion: examples. i hope this doesn’t apply to you!

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