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					MCV 4U1
                               Monkeys, Bananas, and Parabolic Motion

       We can use 2D vectors to analyze the motion of projectiles that are launched into the air. We can
use vectors with our knowledge of derivatives to determine the optimum launch angles as well as
determining horizontal ranges and maximum height reached by objects launched into the air.

In this activity, we will be trying to launch a banana with a canon so that it lands in the hands of a falling
monkey. Strange…but true!

Instructions
1) Visit the webpage
http://www.explorelearning.com/index.cfm?method=cResource.dspResourcesForCourse&CourseID=33
0

2) Launch the Shoot the Monkey Gizmo

3) Try to launch the banana so that the Monkey has a snack on his trip towards the ground!

4) Click on the Show Velocity Vectors box.

Part A
1) Which of the velocity vectors corresponds to the velocity of the falling monkey? How do you know
this?




2) What does the negative value for velocity mean?
                                                                                                         
2) Using the angle and the magnitude of the initial velocity vector ( v i ) for the banana, calculate the v x
    
and v y components of the velocity vector




3) What happens to the x-component of the banana velocity vector after the launch from the cannon?




                                                                                                                1
4) What happens to the y-component of the banana velocity vector after the launch from the cannon?




5) Why does the x-component stay constant and the y-component change?




Part B
When a projectile is launched, the equations for the distance traveled in the horizontal and vertical
directions can be calculated. For the vertical direction, we have:

                                     1
                       h(t )  viy t  gt 2 ,
                                      2
where h is the height of the object above the ground at
         
time t, v iy is the initial y-component of the velocity, and
 
 g is the acceleration from gravity. On earth the
acceleration due to gravity is 9.8 m/s 2 [down].
                                   
For the horizontal direction:     d x  vx t


1) Using the component method, determine the equations for horizontal and vertical position in terms of
                            
the initial velocity vector v i and the launch angle θ. Show your work below.




2) From the equations above, we see that the height h is a function of time t. What kind of function is
h(t)?




                                                                                                          2
3) Using the concepts of calculus, how would you determine the maximum height reached by the
projectile?




4) A banana is launched at an initial velocity of 15 m/s at an angle of 25˚ to the horizontal.

a) Determine the maximum height reached by the projectile. Verify your answer by inputting the
above values into the Gizmo.




b) Using the Gizmo, determine the vertical velocity at the maximum height. At what time does this
occur? How does this time compare to the time calculated in a) above?




                                                                                                    3
c) Determine the equation for the vertical component of velocity using the concepts of calculus. Using
this equation, calculate the velocity of the banana 0.5 s and 1.1 s after launch. Verify your answer by
using the Gizmo.




5) Using the equation for the horizontal distance found in question 1) above, Show that the derivative of
the horizontal distance with respect to time is a constant value if the launch angle and initial velocity do
not change.




6) What does the derivative of the horizontal distance represent?


7) Using your knowledge of trigonometric derivatives, show that the height and horizontal distance are
both maximized if the launch angle is 45˚.




                                                                                                           4
Part C: Practice!

1. A ball is thrown horizontally off a cliff that is 25 m above ground level. Determine

a) the time of flight

b) The horizontal distance the ball lands from the cliff

c) the impact velocity of the ball.




2. A ball is launched from a cliff 55 m above ground level with a velocity of 23 m/s, 36˚ above the
horizontal. Determine

a) the maximum height reached by the ball, relative to the ground.

b) the horizontal distance the ball lands from the cliff.




                                                                                                      5
3. A monkey throws a banana to a chimpanzee as shown. Determine
the velocity and angle the banana must be thrown at if the chimpanzee
catches the banana as it reaches its maximum height. Assume the
chimpanzee is 15 m above where the banana is released and 10 m to
the right of it.




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posted:3/27/2011
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