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Hard Projectile Problem

  • Thread starter roam
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  • #1
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Homework Statement



The diagram below illustrates the ballistic flight of circus daredevil, The Great Zacchini, over three ferris wheels, each 21.2 m high. Zacchini is launched at a speed of 27.6 m/s from the muzzle of the human cannon at an angle of 56.7 ° to the horizontal from a height of 1.6 m above the ground. The net in which he lands is located at the same height above the ground. The Ferris wheels are arranged an equal distance apart, the one closest to the cannon being displaced 24.1 m from the cannon's muzzle. For the purposes of the calculations below assume that air resistance is negligible and treat Zacchini as a point particle.

http://img510.imageshack.us/img510/793/imagebz.jpg [Broken]

(a) By how much does he clear the first Ferris wheel?

(b) If he reached his maximum elevation above the middle Ferris wheel what is his clearance above it?

The Attempt at a Solution



(a)
[tex]v_{ix}=27.6 cos (56.7) = 15.15[/tex]
[tex]v_{iy}=27.6 sin (56.7) = 23[/tex]

[tex]x=vt \Rightarrow 24.1 = 15.15 t[/tex]
[tex]t=1.6[/tex]

[tex]y=v_{iy}t+\frac{1}{2}(9.81)(1.6)^2 = 49.35[/tex]

49.35 is the height it obtains when it is over the first Ferris wheel. Since the heigh of the Ferris wheel is 21.2 we have 49.35-21.2=28.15

But this is wrong because the correct answer has to be 4.69. So what are my mistakes? :confused:

(b) No idea how to get started!
 
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Answers and Replies

  • #2
ehild
Homework Helper
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1,854
[tex]
y=v_{iy}t+\frac{1}{2}(9.81)(1.6)^2 = 49.35
[/tex]

This is wrong. Great Zacchini flies upward, but the acceleration points downward.
More: Note that he started his fly at height of 1.6 m above ground, and the top of the Ferris wheels are at 24.1 m above ground.

ehild
 
  • #3
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11
[tex]
y=v_{iy}t+\frac{1}{2}(9.81)(1.6)^2 = 49.35
[/tex]

This is wrong. Great Zacchini flies upward, but the acceleration points downward.
Did you mean that it should be -9.81? But I thought for the vertical motion, taking down as positive, we have

[tex]y=23 \times 1.6+\frac{1}{2}(9.81)(1.6)^2 = 49.35[/tex]

I'm not sure if I undertand your point very well :shy:

More: Note that he started his fly at height of 1.6 m above ground, and the top of the Ferris wheels are at 24.1 m above ground.

ehild
Oh thanks! So we add up 1.6 to the height of the Ferris wheels 24.1. That is 25.7, but if subtract it from 49.35 we end up with 23.6 which is again not the correct answer.
 
  • #4
rl.bhat
Homework Helper
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You are measuring y in the upward direction and g in the downward direction. So if you take g positive then the y must be negative.
 
  • #5
ehild
Homework Helper
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1,854
Did you mean that it should be -9.81? But I thought for the vertical motion, taking down as positive, we have

[tex]y=23 \times 1.6+\frac{1}{2}(9.81)(1.6)^2 = 49.35[/tex]

I'm not sure if I undertand your point very well :shy:
Just think what happens to Zachhini if your formula is right? Where will he be after 60 s? Will he get back to the ground ever?

ehild
 
  • #6
1,266
11
I see. :rofl:

24.1-1.6 = 22.5 (height of the Ferris). And using -g gives 24.24 so 24.24-19.6=4.69. *correct answer*

So how do I need to approach part (b)? The problem is that I don't know the spacing between the Ferris wheels. That means I can't find the distance traveled from [tex]x=vt[/tex] and therefore I don't know what the time is so it's impossible to find the height at that point!
 
  • #7
rl.bhat
Homework Helper
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When Zaccini reaches the maximum height, vfy is zero.
Using the formula
vfy^2 = viy^2 - 2*g*h find the maximum height.
 
  • #8
ehild
Homework Helper
15,477
1,854
You can find the time when Zacchini reaches at maximum height: it is when his vertical velocity becomes zero, as rl.bhat said. You know that the vertical velocity decreases with time according to the equation

[tex]
v_y=v_{iy}-gt [/tex]

If you know the time, you can calculate the height, as you did with the first wheel.

ehild
 
  • #9
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11
Thanks a lot.

By the way, what is the quickest method for determining the spacing between the Ferris wheels? Can I just find out the difference between the first time and the second time and then use the equation x=vt? Is that accurate?
 
  • #10
rl.bhat
Homework Helper
4,433
7
Thanks a lot.

By the way, what is the quickest method for determining the spacing between the Ferris wheels? Can I just find out the difference between the first time and the second time and then use the equation x=vt? Is that accurate?
Find the time to reach the maximum height.
Find the horizontal distance X between the stating point and the maximum point.
Distance d between the stating point and the first wheel is given.
So distance between first and the second wheel = x - d =....?
 
  • #11
ehild
Homework Helper
15,477
1,854
Thanks a lot.

By the way, what is the quickest method for determining the spacing between the Ferris wheels? Can I just find out the difference between the first time and the second time and then use the equation x=vt? Is that accurate?
It is, but do you need the spacing between the wheels? Use the relation between y and t, the time when the maximum height is reached. Calculate the maximum height. It is just above the middle Ferris wheel, so you can calculate by how much Zacchini clears it.

ehild
 

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