by Richard J. Terwilliger Arrrgh Mate! That Pirate Ship won’t get my booty.

by Richard J. Terwilliger

Arrrgh Mate! That Pirate Ship won’t get my booty.

Luckily I know some physics!

Watch this volley. Arrrgh!

Let me go back and show you how I did this.

First I’ll put back yar ship.

Now lets show the position of the volley for each second of flight.

To make the problem easy, we will ignore any resistance due to air drag.

There are two different directions of motion we are concerned about.

The up-down direction we call the Y-direction.

and the forward direction we call the X-direction.

These two directions are independent of each other.

This means that one direction does NOT directly affect the other direction.

To clearly distinguish between the Y-direction and the X-direction

We will color code the Y-direction RED and the X-direction BLUE.

In the forward direction there is no force either speeding up...

…or slowing down the shot as long as we ignore air resistance.

From this we conclude that the acceleration in the forward direction is zero. d x = V ix t+a x t 2 2 a x =0 a x = zero

If the ball is not accelerating then it must be moving with a constant velocity. d x = V ix t+a x t 2 2 a x =0 V x = constant a x = zero

And if it is moving with a constant velocity then it will cover equal distances in equal amounts of time. d x = V ix t+a x t 2 2 a x =0 V x = constant a x = zero d x = V ix t

We’ll place a ruler at the bottom so we can measure the ball’s forward travel each second. d x = V ix t+a x t 2 2 a x =0 1234567890 V x = constant a x = zero

Next we’ll drop vertical lines that go through the center of each ball’s position. 1234567890 V x = constant a x = zero

As you can see, these lines are equally spaced. 1234567890 V x = constant a x = zero

The ball moved forward one unit distance each second. 1234567890 V x = constant a x = zero

Let’s plot the forward position of the ball each second. 1234567890 V x = constant a x = zero

We’ll pick a forward velocity of the ball to be 10 m/s. 1234567890 V x = constant a x = zero

0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance During the first second the ball travels 10 meters forward. d x = V ix t d x = (10 m)(1s) s d x = 10 m

0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance During the second second the ball travels another 10 meters forward. d x = V ix t d x = (10 m)(1s) s d x = 10 m (2s) 20 m

0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance Each second the ball travels another 10 meters forward. d x = V ix t d x = (10 m)(1s) s d x = 10 m (2s) 20 m ( t )

0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance The distance vs time graph of the ball’s forward motion.. d x = V ix t d x = (10 m) s ( t )

0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance Is a straight line with a positive slope. d x = V ix t d x = (10 m) s ( t )

0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance The slope represents the ball’s VELOCITY in the forward direction. d x = V ix t d x = (10 m)(t) s  d x  t V x =

 d x  t V x = 0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance The slope represents the ball’s VELOCITY in the forward direction. d x = V ix t = d 2 – d 1 t 2 – t 1 80 m – 20 m 8 s – 2 s = = 10 m s

0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance As shown by our graph, the ship was 100 meters away. 100 meters

0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance We can also plot a velocity vs. time graph representing the forward motion of the ball. 0 24 6 8 10 time (seconds) 20 10 0 Forward velocity vs. time (V x ) (m/s) velocity

0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance Notice the velocity in the forward direction is constant. 0 24 6 8 10 time (seconds) 20 10 0 Forward velocity vs. time (V x ) (m/s) velocity

0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance The slope is ZERO and represents the acceleration in the forward direction. 0 24 6 8 10 time (seconds) 20 10 0 Forward velocity vs. time (V x ) (m/s) velocity

0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance The area of the velocity-time graph represents the forward distance traveled each second by the ball. 0 24 6 8 10 time (seconds) 20 10 0 Forward velocity vs. time (V x ) (m/s) velocity

0 24 6 8 10 100 80 60 40 20 0 Forward travel vs. time time (seconds) (meters) (d x ) distance Summarizing the forward direction: 0 24 6 8 10 time (seconds) 20 10 0 Forward velocity vs. time V x = constant a x = zero (V x ) (m/s) velocity

Next we’ll look at what is taking place in the Y-direction or the up-down direction. V x = constant a x = zero

Again, for every second of travel, we’ll draw lines through the centers of each ball. V x = constant a x = zero

Again, for every second of travel, we’ll draw lines through the centers of each ball. 1 2 3 4 5 6 7 8 9 0 V x = constant a x = zero

Notice that on the way up to the peak, the distance the ball travels each second decreases. 1 2 3 4 5 6 7 8 9 0 V x = constant a x = zero

1 2 3 4 5 6 7 8 9 0 V x = constant a x = zero Therefore the ball must be slowing down in the up direction or accelerating in the direction.

So, on the way up the ball is accelerating 1 2 3 4 5 6 7 8 9 0 V x = constant a x = zero a y a y a y a y a y

And on the way back down, the distance the ball travels each second increases. 1 2 3 4 5 6 7 8 9 0 V x = constant a x = zero

This means the ball is speeding up on the way back down. 1 2 3 4 5 6 7 8 9 0 V x = constant a x = zero

V x = constant a x = zero Again, accelerating in the direction. a y a y a y a y a y a y

If we define down as V x = constant a x = zero

The ball has a negative acceleration on the way up and the way back down. V x = constant a x = zero a y a y a y a y a y a y a y a y a y

As it turns out, the acceleration of the ball in the up-down direction is V x = constant a x = zero a y a y a y a y a y a y a y a y a y a y = - 9.81 m/s 2

V x = constant a x = zero a y = - 9.81 m/s 2 Let’s we’ll plot the displacement, velocity and acceleration of the ball in the Y-direction.

0 24 6 8 10 time (seconds) Vertical Acceleration vs. time (A y ) (m/s 2 ) Acceleration 10 5 0 -5 -10 The Vertical Acceleration vs. time graph is a straight line parallel to the time axis with a value of –9.81 m/s 2.

Next we will plot the Vertical Velocity vs. time

To determine the Vertical Velocity we use the equation:

and the acceleration in the Y direction is We know the projectile was in the air for 10 seconds

Knowing the acceleration in the Y direction is constant we can conclude that the time to the peak was one half the total time.

We also know that the velocity at the peak in the Y-direction is zero.

Now we have enough information to solve for the initial velocity in the Y-direction

First we’ll rearrange the equation solving for the initial velocity.

Next we’ll substitute in our values. Don’t leave out the units! = 0 m/s – (-9.81 m/s 2 )(5s) Peak = 0

Finally solving = 0 m/s – (-9.81 m/s 2 )(5s) = +49.05 m/s

So the projectile was shot with an initial vertical velocity of +49.05 m/s = 0 m/s – (-9.81 m/s 2 )(5s) = +49.05 m/s

Now we’ll solve for the velocity in the Y direction for every second of flight and record it in a table.

Next we’ll solve for the velocity in the Y direction for every second of flight and record it in a table. = +49.05 m/s + (-9.81 m/s 2 )( ) = +39.24 m/s Velocity (m/s) Time (s) 0 +49.05 1+39.24 1s

= +49.05 m/s + (-9.81 m/s 2 )( ) = +39.24 m/s Velocity (m/s) Time (s) 0 +49.05 1+39.24 1s After going through a few calculations and you get the knack of it, feel free to click on me to skip ahead.

Now we’ll solve for the velocity in the Y direction for every second of flight and record it in a table. = +49.05 m/s + (-9.81 m/s 2 )( ) = +29.43 m/s Velocity (m/s) Time (s) 0 +49.05 1+39.24 2s +29.43 2

Now we’ll solve for the velocity in the Y direction for every second of flight and record it in a table. = +49.05 m/s + (-9.81 m/s 2 )( ) = +19.62 m/s Velocity (m/s) Time (s) 0 +49.05 1+39.24 3s +29.43 2 3 +19.62

Now we’ll solve for the velocity in the Y direction for every second of flight and record it in a table. = +49.05 m/s + (-9.81 m/s 2 )( ) = +9.81 m/s Velocity (m/s) Time (s) 0 +49.05 1+39.24 4s +29.43 2 3 +19.62 4 +9.81

Now we’ll solve for the velocity in the Y direction for every second of flight and record it in a table. = +49.05 m/s + (-9.81 m/s 2 )( ) = 0 m/s Velocity (m/s) Time (s) 0 +49.05 1+39.24 5s +29.43 2 3 +19.62 4 +9.81 5 0

Do you notice that the velocity is decreasing by 9.81 m/s each second? = +49.05 m/s + (-9.81 m/s 2 )( ) = 0 m/s Velocity (m/s) Time (s) 0 +49.05 1+39.24 5s +29.43 2 3 +19.62 4 +9.81 5 0

Let’s continue to solve for the velocity up to a time of flight of 10 seconds. = +49.05 m/s + (-9.81 m/s 2 )( ) = -9.81 m/s Velocity (m/s) Time (s) 0 +49.05 1+39.24 6s +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81

Let’s continue to solve for the velocity up to a time of flight of 10 seconds. = +49.05 m/s + (-9.81 m/s 2 )( ) = -19.62 m/s Velocity (m/s) Time (s) 0 +49.05 1+39.24 7s +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62

Let’s continue to solve for the velocity up to a time of flight of 10 seconds. = +49.05 m/s + (-9.81 m/s 2 )( ) = -29.43 m/s Velocity (m/s) Time (s) 0 +49.05 1+39.24 8s +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43

Let’s continue to solve for the velocity up to a time of flight of 10 seconds. = +49.05 m/s + (-9.81 m/s 2 )( ) = -39.24 m/s Velocity (m/s) Time (s) 0 +49.05 1+39.24 9s +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24

Let’s continue to solve for the velocity up to a time of flight of 10 seconds. = +49.05 m/s + (-9.81 m/s 2 )( ) = -49.05 m/s Velocity (m/s) Time (s) 0 +49.05 1+39.24 10s +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24 10 -49.05

Notice that the initial speed is equal to the final speed. Velocity (m/s) Time (s) 0 +49.05 1+39.24 +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24 10 -49.05

Now we have the information needed to plot a vertical velocity vs. time graph. Velocity (m/s) Time (s) 0 +49.05 1+39.24 +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24 10 -49.05

Velocity (m/s) Time (s) 0 +49.05 1+39.24 +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24 10 -49.05 0 24 6 8 10 time (seconds) Vertical Velocity vs. time (V y ) (m/s) Velocity 0 -40 -20 +20 +40 Plotting the vertical velocity vs. time gives us a straight line with a negative slope.

Velocity (m/s) Time (s) 0 +49.05 1+39.24 +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24 10 -49.05 0 24 6 8 10 time (seconds) Vertical Velocity vs. time (V y ) (m/s) Velocity 0 -40 -20 +20 +40 The constant slope shows us the velocity is changing at a constant rate.

Velocity (m/s) Time (s) 0 +49.05 1+39.24 +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24 10 -49.05 0 24 6 8 10 time (seconds) Vertical Velocity vs. time (V y ) (m/s) Velocity 0 -40 -20 +20 +40 The slope represents the vertical acceleration

Velocity (m/s) Time (s) 0 +49.05 1+39.24 +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24 10 -49.05 0 24 6 8 10 time (seconds) Vertical Velocity vs. time (V y ) (m/s) Velocity 0 -40 -20 +20 +40 At any point on this line the projectile has an acceleration of

Velocity (m/s) Time (s) 0 +49.05 1+39.24 +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24 10 -49.05 0 24 6 8 10 time (seconds) Vertical Velocity vs. time (V y ) (m/s) Velocity 0 -40 -20 +20 +40 So, at the peak the projectile comes to a stop but still is accelerating.

Velocity (m/s) Time (s) 0 +49.05 1+39.24 +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24 10 -49.05 0 24 6 8 10 time (seconds) Vertical Velocity vs. time (V y ) (m/s) Velocity 0 -40 -20 +20 +40 The area under the line again represents the distance traveled but this time in the Y-direction.

Velocity (m/s) Time (s) 0 +49.05 1+39.24 +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24 10 -49.05 0 24 6 8 10 time (seconds) Vertical Velocity vs. time (V y ) (m/s) Velocity 0 -40 -20 +20 +40 The area shaded above represents the distance traveled to the peak.

Velocity (m/s) Time (s) 0 +49.05 1+39.24 +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24 10 -49.05 0 24 6 8 10 time (seconds) Vertical Velocity vs. time (V y ) (m/s) Velocity 0 -40 -20 +20 +40 Notice that the distance traveled per unit time decreases as the ball gets closer to the peak.

Velocity (m/s) Time (s) 0 +49.05 1+39.24 +29.43 2 3 +19.62 4 +9.81 5 0 6-9.81 7-19.62 8-29.43 9-39.24 10 -49.05 0 24 6 8 10 time (seconds) Vertical Velocity vs. time (V y ) (m/s) Velocity 0 -40 -20 +20 +40 After the peak, the ball covers more distance per unit time as it falls.

Next we’ll plot a graph of displacement vs. time.

Again, we need to calculate the displacement for each second and record our results in a table. Displacement (m) Time (s)

We’ll use the displacement equation: Displacement (m) Time (s)

And substitute for time. Displacement (m) Time (s) = +49.05 m/s ( t ) + (-9.81 m/s 2 )( t ) 2 2

= +49.05 m/s ( ) + (-9.81 m/s 2 )( ) 2 = 0 m 00 0s Displacement (m) Time (s) And substitute for time. 2

= +49.05 m/s ( ) + (-9.81 m/s 2 )( ) 2 = +44.145 m 00 1s Displacement (m) Time (s) And substitute for time. 1+44.145 2

= +49.05 m/s ( ) + (-9.81 m/s 2 )( ) 2 = +78.480 m 00 2s Displacement (m) Time (s) And substitute for time. 1+44.145 2+78.480 2

= +49.05 m/s ( ) + (-9.81 m/s 2 )( ) 2 = +103.005 m 00 3s Displacement (m) Time (s) And substitute for time. 1+44.145 2+78.480 2 3 +103.005

= +49.05 m/s ( ) + (-9.81 m/s 2 )( ) 2 = +117.720 m 00 4s Displacement (m) Time (s) And substitute for time. 1+44.145 2+78.480 2 3 +103.005 4+117.720

= +49.05 m/s ( ) + (-9.81 m/s 2 )( ) 2 = +122.625 m 00 5s Displacement (m) Time (s) And substitute for time. 1+44.145 2+78.480 2 3 +103.005 4+117.720 5 +122.625

= +49.05 m/s ( ) + (-9.81 m/s 2 )( ) 2 = +117.720 m 00 6s Displacement (m) Time (s) And substitute for time. 1+44.145 2+78.480 2 3 +103.005 4+117.720 5+122.625 6 +117.720

= +49.05 m/s ( ) + (-9.81 m/s 2 )( ) 2 = +103.005 m 00 7s Displacement (m) Time (s) And substitute for time. 1+44.145 2+78.480 2 3 +103.005 4+117.720 5+122.625 6 +117.720 7 +103.005

= +49.05 m/s ( ) + (-9.81 m/s 2 )( ) 2 = +78.480 m 00 8s Displacement (m) Time (s) And substitute for time. 1+44.145 2+78.480 2 3 +103.005 4+117.720 5+122.625 6 +117.720 7 +103.005 8+78.480

= +49.05 m/s ( ) + (-9.81 m/s 2 )( ) 2 = +44.145 m 00 9s Displacement (m) Time (s) And substitute for time. 1+44.145 2+78.480 2 3 +103.005 4+117.720 5+122.625 6 +117.720 7 +103.005 8+78.480 9+44.145

= +49.05 m/s ( ) + (-9.81 m/s 2 )( ) 2 = 0 m 00 10s Displacement (m) Time (s) And substitute for time. 1+44.145 2+78.480 2 3 +103.005 4+117.720 5+122.625 6 +117.720 7 +103.005 8+78.480 9+44.145 100

00 Displacement (m) Time (s) Now we have the data needed to plot our vertical displacement vs. time graph. 1+44.145 2+78.480 3 +103.005 4+117.720 5+122.625 6 +117.720 7 +103.005 8+78.480 9+44.145 100 0 24 6 8 time (seconds) Vertical Displacement vs. time (D y ) (m) Displacement 0 120 80 40

00 Displacement (m) Time (s) Notice the graph looks like the actual path of the projectile if viewed from the side. 1+44.145 2+78.480 3 +103.005 4+117.720 5+122.625 6 +117.720 7 +103.005 8+78.480 9+44.145 100 0 24 6 8 time (seconds) Vertical Displacement vs. time (D y ) (m) Displacement 0 120 80 40

Let’s REVIEW

t d In the X-direction, the plot of a displacement vs. time was a straight line with a positive slope.

t d t v The plot of a velocity vs. time is a straight line parallel to the time axis. Therefore the slope is zero.

t d t v ta The plot of an acceleration vs. time is a straight line parallel to the time axis and the acceleration has a value of zero.

t d t v ta t d In the Y direction, a plot of the displacement vs. time is a parabola.

t d t v ta t d tv A plot of velocity vs. time is straight line with a negative slope.

t d t v ta t d tv ta A finally, the plot of the acceleration vs. time is straight line parallel to the time axis with a value of -9.81 m/s 2.

V ix = V iy = Knowing the initial velocity in the forward direction and the initial velocity, in the upward direction

and at what angle. V ix = V iy = We can determine how fast the projectile was shot 

We refer to this as the V ix = V iy = 

Let’s zoom in on the vectors. V ix = V iy =

We’ll label the initial velocity as Vi V ix = V iy =

Next we’ll move the Y- velocity vector so we have a right triangle that is easier to visual. V ix = V iy =

V ix = V iy = Next we’ll move the Y- velocity vector so we have a right triangle that is easier to visual.

V ix = V iy = Label the X and Y velocities.

V iy = V ix =

Using the Pythagorean Theorem we can solve for the V iy = V ix = of the initial velocity. c 2 =a 2 +b 2 V i 2 =V iy 2 +V ix 2 V i = V iy 2 +V ix 2

Using the Pythagorean Theorem we can solve for the V iy = V ix = of the initial velocity. V i = V iy 2 +V ix 2 V i = (49 m/s) 2 +(10 m/s) 2 V i = 50 m/s

We can also solve for the direction of the velocity using V iy = V ix =  opposite adjacent tan  =

We can also solve for the direction of the velocity using V iy = V ix =  tan  = Vix Viy

We can also solve for the direction of the velocity using V iy = V ix =   = = 10 m/s (tan -1 ) 49.05 m/s  = = 78.5 o tan  = Vix Viy (tan -1 )

We can also solve for the direction of the velocity using V iy = V ix =  = = 10 m/s (tan -1 ) 49.05 m/s  = = 78.5 o tan  = Vix Viy (tan -1 ) 78.5 o

The initial velocity of the projectile was V iy = V ix =  = = 10 m/s (tan -1 ) 49.05 m/s  = = 78.5 o tan  = Vix Viy (tan -1 ) 78.5 o @

by Richard J. Terwilliger Arrrgh Mate! That Pirate Ship won’t get my booty.

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by Richard J. Terwilliger Arrrgh Mate! That Pirate Ship won’t get my booty.

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