Dynamics Problem

8/18/2019 Dynamics Problem

1/16

Projectile Motion

g~ ax = dvx/dt = 0

ay = dvy/dt = -g

vx = vx,o

vy = vy,o - gt

x = vx,o t + xo

y = yo + vy,o t - gt2

2Friday, November 11, 2011


2/16

Projectile Motion

g~ ax = dvx/dt = 0

ay = dvy/dt = -g

vx = vx,o

vy = vy,o - gt

x = vx,o t + xo




3/16

Projectile Motion

g~ ax = dvx/dt = 0

ay = dvy/dt = -g

vx = vx,o

vy = vy,o - gt

x’

y’

x = vx,o t + xo




4/16

Projectile Motion

Friday, November 11, 2011


5/16

ExampleGiven: Given vo and #, find the equationthat defines y as a function of x.

Use vx = vo cos # and vy = vo sin #

x = (vo cos #)t or t =x

vo cos !

y = (vo sin #) t – $ g (t)2

Eliminate t :

y = (vo sin !) { } { }2x g x

vo cos ! 2 vo cos ! –



6/16

Example 2

Projectile is fired with vA=150 m/s at point A.

The horizontal distance it travels (R) and thetime in the air.



7/16

Example 2

Establish a fixed x, y coordinate system (in this solution, theorigin of the coordinate system is placed at A). Apply thekinematic relations in x- and y-directions.

Projectile is fired with vA=150 m/s at point A.

The horizontal distance it travels (R) and thetime in the air.



8/16

Example 2

1) Place the coordinate system at point A.

Equation for horizontal motion.

xB = xA + vAx tf where xB = R, xA = 0, vAx = 150 (4/5) m/s

R = 120 tf Time of flight: tf



9/16

Example 2




R = 120 tf

2) Now write a vertical motion equation. Use the distance equation.

yB = yA + vAy tf – 0.5 g tf 2

where yB = – 150, yA = 0, and vAy = 150(3/5) m/s

We get the following equation: –150 = 90 tf + 0.5 (– 9.81) tf 2

Time of flight: tf



10/16

Example 2




R = 120 tf

2) Now write a vertical motion equation. Use the distance equation.

yB = yA + vAy tf – 0.5 g tf 2

where yB = – 150, yA = 0, and vAy = 150(3/5) m/s

We get the following equation: –150 = 90 tf + 0.5 (– 9.81) tf 2

Solving for tf first, tf = 19.89 s.

Then, R = 120 tf = 120 (19.89) = 2387 m

Time of flight: tf



11/16

Example 3

A skier leaves the ski jump ramp at "A

= 25o and hits the slope at B. Find theSkier’s initial speed, va.

Motion in x-direction:

Using xB = xA + vox(tAB) => (4/5)100 = 0 + vA (cos 25) tAB

=tAB=80

vA

(cos 25°)

88.27

vA

x

y



12/16

Example 3

Motion in y-direction:

Using yB = yA + voy(tAB) – $ g(tAB)2

– 64 = 0 + vA(sin 25°) { }88.27

vA – ! (9.81) { }2

88.27

vA

vA = 19.42 m/s



13/16

Do it in your head1. In a projectile motion problem, what is the maximum number of unknownsthat can be solved?

A) 1 B) 2

C) 3 D) 4



14/16

Do it in your head1. In a projectile motion problem, what is the maximum number of unknownsthat can be solved?

A) 1 B) 2

C) 3 D) 4

2. The time of flight of a projectile, fired over level ground, with initialvelocity Vo at angle #, is equal to?

A) (vo sin ")/g B) (2vo sin ")/g

C) (vo cos ")/g D) (2vo cos ")/g



15/16

Do it in your head

3. A projectile is given an initial velocity

vo at an angle # above the horizontal.

The velocity of the projectile when it

hits the slope is ____________ the

initial velocity vo.

A) less than B) equal to

C) greater than D) None of the above.



16/16

Do it in your head

3. A projectile is given an initial velocity

vo at an angle # above the horizontal.

The velocity of the projectile when it

hits the slope is ____________ the

initial velocity vo.

A) less than B) equal to

C) greater than D) None of the above.

4. A particle has an initial velocity vo at angle " with respect to the horizontal. The

maximum height it can reach is when

A) " = 30° B) " = 45°

C) " = 60° D) " = 90°

Friday November 11 2011

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Dynamics Problem