If you're interested in previous attempts to shoot a ballistic (non-powered) projectile into space, check out US ARMY project HARP that achieved a near space altitude of 180 km.

Note: ICBM (inter-continental ballistic missiles) that we see on TV are launched via missiles to their highest altitude, and then they fall back to space in 'ballistic' trajectories. Although those trajectories can be altered by adjustments to air flow over the projectile, they are not themselves under power as they fall)

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LEARNING OBJECTIVE:

I will be able to determine the time at which a projectile reaches it's highest point after today's class.

I will be able to determine the range at which a projectile travels after today's class.

I will be able to sketch the vectors for the x component and y component of velocity for a projectile after today's class.

WORDS FOR TODAY:

• DISPLACEMENT (distance & direction)
• Distance
• VELOCITY (speed & direction or change in displacement/change in time)
• Speed (change in distance/change in time)
• INSTANTANEOUS VELOCITY (dx/dt)
• ACCELERATION (change in velocity/change in time, dv/dt)
• Derivative (the instantaneous slope of a line/curve at a given point)
• Integral (the area 'beneath' a line or curve)

CALENDAR: Test on Thursday, Labs due on Friday

WORK O' THE DAY: 

What were two reasons for having you work the battleship problem?

Let's try another working example:

In previous examples we tossed or shot objects straight up and let them fall straight back down.... clearly 1 dim motion.

Now let's consider the more likely situation of an object traveling in projectile (parabolic) motion:

Instead of throwing the rock straight up, imagine you were standing on a 45 meter tall building and you tossed a rock at a 30.0 degree angle to the horizontal with an initial velocity of 20.0 m/s. How far away from the base of the building does the rock fall if we disregard air friction, wind, humidity, temperature and the other usual considerations?

1) Write a sketch of that situation that shows ALL initial conditions on the sketch (be sure to include the x and y components of the velocity of that object)

2) Write a short qualitative analysis for JUST THE Y motion of that object

3) Write a short qualitative analysis for JUST THE X motion of that object

Hopefully you have something that looks like this:

SOLUTION:

We absolutely, positively MUST break this down into x and y components of velocity (besides, I asked yout to analyze JUST the y motion).

So, we realize that we can analyze this problem in terms of motion in one dimension (y) and then use that information (if we need to) to analyze that object's motion in the x direction.

So... we know the object goes up for a certain length of time until the acceleration due to gravity stops that initial motion, and then the object comes right back down again past the top of the building, accelerating as it goes (due to gravity) and then comes to rest 45 meters below the top of the building.

So let's list our initial conditions:

v_i,y = v_isinθ=(20. m/s)(sin30.0) = 10.0 m/s

v_i,x = v_icosθ=(20. m/s)(cos30.0) = 17.3 m/s

v_f,y = 0 m/s

y_i = 45 m

y_f = 0 m

a = -9.81 m/s²

launch angle = 30.0 degrees to the horizontal

You have a couple of different approaches here:

Approach #1)

Step 1) Find the time for the object to rise until it is stopped by gravity

Step 2) Find out how long it takes to fall back to earth

Approach #2)

Do it all at once using one of our hard working motion equations:

y_f = y_i + v_i,yt + 1/2at²

So... inserting our initial conditions:

0m = 45m + (10m/s)t + (.5)(-9.8m/s²)t²

=

0m = 45m + (10m/s)t + (-4.9m/s²)t²

Solving for t by pasting into good ol' wolfram alpha

We get 4.22 sec! (siggie figgies)

Please notice my approach is slightly different then the book approach... please compare both and convince yourself that both are accurate!

Recall that the displacement downrange is simply:

(v_ix)(t)

Sooooo.... recall from our initial conditions that:

v_ix = v_icosθ=(20. m/s)(cos30.0) = 17.3 m/s

(17.3 m/s)(4.22 s) = 73.09 m = 73 m s.f.

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Take a brief look at the Long Jump problem on page 87. Notice that they use the 'range equation' for that.

That equation is very much a plug-and-chug sorta thing. Although that can be satisfying *ping* to fill in the slots and do the math and get the right answer, we will rarely (if ever) use it. We will derive what we need from our "regular" equations of motion.

Just for practice, work with your group and see if you can develop an approach to solving those questions WITHOUT using the range equation.

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Let's use our favorite wrist-rocket again for another example:

You're standing on top of a building 253 meters tall. You take your trusty-dustry wrist-rocket and shoot a ball bearing at an angle 49.55 degrees to the horizontal at a initial velocity of 21.23 m/s. The projectile lands on top of another building that is 195.5 meters in height. How far away is the other building?

• Sketch that situation.

Talk with your group to:

• Determine how that problem is new and different from a previous ballistic motion problems.
• Plan a new approach.
• Do it!
• Answers are below

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HOMEWORK:

• Please review worked problems: 4.5 on page 90
• Please take a swing at O problem #6 on page 100
• And then C problem: #5 on page 101
• And problems #21 and 25