Verticle Projectile Problem

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In summary: Is there another way to let people know that someone has solved it?In summary, Phil launches a potato from a cliff and observes that it rises to a maximum height of 56.5 meters and hits the water in a total time of 12.31 seconds.
  • #1
smzee27
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Homework Statement



Note: This is supposed to be simplistic so we're only concerned with the vertical movement. All horizontal factors are completely ignored and should not be included in the solutions or the procedures to get to the solutions.

Phil wants to test his new potato launcher at the edge of a cliff overlooking the ocean. He launches his potato from the edge of the cliff and observes that it rises to a maximum height of 56.5 meters and hits the water in a total time of 12.31 seconds. Ignoring the effects of air resistance, answer the following:

What is the initial velocity v(0y) of the potato at the launch point.

Homework Equations



[tex]g = \frac {v_y - v_0_y} {t}[/tex]
[tex]v_y = v_0_y + g t[/tex]
[tex]y = y_0 + v_0_y t + \frac{1}{2} g t^2[/tex]
[tex]v_y^2 - v_0_y^2 = 2 g \Delta y[/tex]
[tex]\Delta y = (\frac {v_y + v_0_y} {2}) t[/tex]
[tex]t = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a} [/tex]

(g= 9.80665 OR g=-9.80665, I'm not sure which applies when.)

The Attempt at a Solution



I tried to use multiple formulas but none of them seem to work right. I only have two pieces of information here and I don't know how to use them in place of any variable. The time given is total. The distance given is only the displacement up to the highest point.

I'm completely stumped. Is there just not enough information? Or maybe I'm over thinking the whole thing and the solution is right there? Either way, I just can't even begin to understand this. I'm not particularly looking for an answer, I just need to know how I'm supposed to do this. There are more questions after this one, but if I can do this, I'll probably be able to do the rest too.
I copied down the problem here exactly as it on my paper. And I reread the problem and this post multiple times to ensure there are no errors. Also, here is a link to a scan of the original problem and diagram as it is on the paper.

EDIT: As I wrote this, I realized something I didn't notice before and have started to work on the problem again. But I would still appreciate your help. Thanks in advance.

http://s100.photobucket.com/albums/m32/smzee27/?action=view&current=Untitled.jpg

I would really appreciate any help on this. I've done these kinds of problems before, but this particular one has me stumped. Although everyone else in the class seems to have gotten through this one, albeit with some difficulty.
 
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  • #2
Energy conservation approach is better here, just equate KE at initial stage to change in PE at final stage (decide what to take as final state).

Alternatively
Newton's way : use your 4th equation.
 
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  • #3
You can do this problem just using the 2nd and 3rd of your listed equations.

Use the 2nd to elminate V0y from the 3rd. Then solve for the time to the top of flight, using the given height. Once you have the time to the top, you know the time to return to the starting point. Now use that time and your knowledge of the velocity to solve the 2nd equation for the initial velocity.
 
  • #4
Sourabh N said:
Energy conservation approach is better here, just equate KE at initial stage to change in PE at final stage (decide what to take as final state).

Alternatively
Newton's way : use your 4th equation.

The class is not at the point yet where we can use something like that; your method seems too advanced right now. Thanks for your reply though.

Integral said:
You can do this problem just using the 2nd and 3rd of your listed equations.

Use the 2nd to elminate V0y from the 3rd. Then solve for the time to the top of flight, using the given height. Once you have the time to the top, you know the time to return to the starting point. Now use that time and your knowledge of the velocity to solve the 2nd equation for the initial velocity.

Assuming I did that first part right, I get:

[tex] y = y_0 + v_y - gt + \frac {1}{2} gt^2 [/tex]

Okay, so [tex] y = 56.5 m [/tex], and [tex]y_0 = 0[/tex]? Also, how would I get [tex]v_y[/tex]? If [tex]v_y[/tex] is supposed to be the average velocity for the whole thing, I only know the time for the whole motion, not the distance.
 
  • #5
Velocity at the peak of the projection is 0. You have the height of 56.5m.

EDIT:Read post below! no need to find time first.
 
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  • #6
Thanks for your help everyone.

By the way, I don't see a "Mark solved" option under Thread tools.
 
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1. What is a vertical projectile problem?

A vertical projectile problem is a physics problem that involves an object being launched or thrown straight up or straight down. The object is affected by only the force of gravity, and the goal is usually to determine its height, velocity, or time in the air.

2. What are the key equations for solving a vertical projectile problem?

The key equations for solving a vertical projectile problem are the kinematic equations:
- v = u + at (final velocity = initial velocity + acceleration x time)
- s = ut + 1/2at^2 (displacement = initial velocity x time + 1/2 x acceleration x time squared)
- v^2 = u^2 + 2as (final velocity squared = initial velocity squared + 2 x acceleration x displacement)

3. How does air resistance affect a vertical projectile?

Air resistance or drag will typically decrease the vertical velocity of a projectile, making it fall slower than it would in a vacuum. This effect is more pronounced for objects with large surface areas, such as feathers or parachutes. In most basic vertical projectile problems, air resistance is assumed to be negligible.

4. Can a vertical projectile reach a height of zero?

Yes, a vertical projectile can reach a height of zero if it is launched straight up and returns to its initial position. This is known as maximum height and is calculated by setting the final height to zero in the displacement equation:
0 = ut + 1/2at^2
t = 0 or t = 2u/a (assuming u is positive for upward motion)

5. How does the angle of launch affect the height and distance of a vertical projectile?

The angle of launch affects the height and distance of a vertical projectile by changing the initial velocity in the x and y directions. A higher angle will result in a longer distance traveled but a lower maximum height. A lower angle will result in a shorter distance traveled but a higher maximum height. The optimal angle for maximum distance is 45 degrees.

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