Kinematics in Two-Dimensions - Explanation?

In summary, the question asks for the vertical height between two climbers when a first aid kit is thrown from one to the other. The initial velocity of the kit is 11 m/s at an angle of 65 degrees above the horizontal. Using the kinematics equations and the known data, it can be calculated that the vertical height between the two climbers is 5.07m. The mistake made by the person asking the question was using the incorrect initial velocity of 11 m/s, when it should have been calculated using SOH CAH TOA to find the vertical component of the velocity.
  • #1
RyanJF
16
0

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.
 
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  • #2
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
I thought I already explained that that wasn't working...

And my class was never given the "three equations of motion". =\
 
  • #4
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
yup, i should have remembered to point out exactly what his mistake was.
 

1. What is kinematics in two-dimensions?

Kinematics in two-dimensions is the study of the motion of objects in two-dimensional space, taking into account both the direction and magnitude of movement. It involves analyzing the position, velocity, and acceleration of an object over time, without considering the forces causing the movement.

2. How is kinematics in two-dimensions different from one-dimension?

Kinematics in one-dimension only considers the motion of objects in a straight line, while kinematics in two-dimensions takes into account movement in both the x and y directions. This allows for a more comprehensive understanding of an object's motion and can be applied to more complex scenarios.

3. What is the importance of understanding kinematics in two-dimensions?

Understanding kinematics in two-dimensions is essential for analyzing and predicting the motion of objects in real-world scenarios. It is the foundation of many other physics principles, such as projectile motion and circular motion, and is crucial in fields such as engineering, robotics, and astronomy.

4. What are the key equations used in kinematics in two-dimensions?

The key equations used in kinematics in two-dimensions are the equations for position, velocity, and acceleration in both the x and y directions. These include equations such as x = x0 + v0t + 1/2at2 for position, v = v0 + at for velocity, and a = Δv/Δt for acceleration.

5. How can kinematics in two-dimensions be applied in real life?

Kinematics in two-dimensions can be applied in real life in various ways, such as predicting the trajectory of a projectile, designing roller coasters, or calculating the motion of planets in the solar system. It is also used in sports, such as analyzing the movement of a soccer ball during a free kick or a baseball during a pitch.

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