Kinematics in Two-Dimensions - Explanation?

  • Thread starter RyanJF
  • Start date
  • #1
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Homework Statement



"A rock climber throws a small first aid kid to another climber who is higher up the mountain. The initial velocity of the kit is 11 m/s at an angle of 65 degrees above the horizontal. At the instant when the kit is caught, it is traveling horizontally, so its vertical speed is zero. What is the vertical height between the two climbers?"

Known data:

Voy = 11 m/s
Vy = 0 m/s
Ay = -9.8 m/s²

Homework Equations



Kinematics equations

The Attempt at a Solution



Alright, so I know the answer is 5.07m, but I have no idea how to get it.

If I use the equation Vy = Voy + AyT and solve for time, I get T = Vy - Voy / A. Plugging in the numbers for the variables, I get an answer of 1.122 seconds = -11 / -9.8. If I plug the value of 1.122 into an equation for y (displacement), I consistently end up with an answer that's off by over 1m.

Here's what I using:

y = 1/2 (Voy + Vy)t

so

6.171 = 1/2 (11)1.122

Needless to say, that answer is 1.01 m from the actual answer of 5.07 m.

I've been stuck for the better part of an hour, trying to figure out how I can solve this equation on my own. I really enjoy physics, and have considered majoring in it (I'm not the best at figuring mathematical things out on my own, but once I get it, I can apply concepts to virtually anything), but not being able to figure out seemingly simple things like this really throws me off.

Using Google and Bing, I've searched both on here and on other websites for an explanation to this problem, yet everything uses equations that are not at all familiar to me. My book and my teacher have only provided the basic equations of kinematics (e.g., V² = Vo² + 2ax), so all of these other bizarre things completely trip me up.

The equation I finally used to solve the problem, also found online, and was not something I learned in class.

It was:

y = Voy²sin(a)/(2g), with G standing for "gravity", or, more correctly, the acceleration of gravity.

I wouldn't have a problem with this sort of thing if we were given equations like this, but unfortunately, we're not.

I'd very much appreciate if someone were to explain how to solve this problem to me, preferably with an even mix of words and math, rather than all one or the other.

Thanks.
 

Answers and Replies

  • #2
424
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v=0
u=11*sin(65)
g=9.8

use one of the kinematics equations and get your answer...(the 3 equations of motion that is.)
 
  • #3
16
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I thought I already explained that that wasn't working...

And my class was never given the "three equations of motion". =\
 
  • #4
1
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Voy = 11 m/s is incorrect. You were given the velocity of the resultant: 11m/s at 65degrees above the horizontal

Using SOH CAH TOA you can get Voy.
sin (65) = Opposite/11.2m/s
sin (65) * 11.2m/s = Opposite
Opposite = 10.151m/s
Voy = 10.151m/s

You should be able to get the rest, just remember to read the question carefully and draw a picture of the situation, the vectors (right angle triangles of the vertical and horizontal component, and the resultant of those components) are extremely important to draw.

Just as a note The legend said this exact thing, but didn't explain it in any detail.
 
  • #5
424
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yup, i should have remembered to point out exactly what his mistake was.
 

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