Another Two Dimensional motion question

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Homework Help Overview

The discussion revolves around two-dimensional motion problems, specifically focusing on projectile motion. The original poster presents a scenario involving a baseball throw and later introduces a swimming scenario involving horizontal motion off a diving board.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the decomposition of velocity vectors into horizontal and vertical components, questioning the use of trigonometric functions for this purpose. There are attempts to clarify the relationship between initial and final velocities in projectile motion, as well as the effects of gravity on vertical motion.

Discussion Status

Participants are actively engaging with the problems, offering suggestions for using trigonometric functions to find components of velocity. There is a mix of interpretations regarding the application of equations and the roles of horizontal and vertical components in the context of projectile motion. Some guidance has been provided, but no consensus has been reached on all aspects of the problems.

Contextual Notes

Assumptions include neglecting air resistance in the baseball throw scenario and the need for clarity on the initial conditions and components of motion in both scenarios. There is also a mention of confusion regarding the application of kinematic equations and the use of trigonometry in the absence of angles in the swimming scenario.

  • #31
v = d/t, you've got it the wrong way around.

And no, x = 0. But initial velocity is zero.
 
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  • #32
joshMp how would you use trig if you only have one number and no degree?
 
  • #33
You have the x component, which was given to you. Find the y component using a kinematic equation. You now have the x component and the y component of the final velocity vector. Vf^2= Vx^2 + Vy^2
 
  • #34
JoshMP said:
"so would the final velocity be 3.60 m/s"

That was the question I was answering earlier when I said to use trig. You need to know the y-component to calculate the final velocity. I was not talking about finding t.

JoshMP said:
^ You need the final velocity to use that equation. Find the time to reach horizontal displacement, then use that time to find the vertical displacement.

My apologies, I thought the initial post (the one I've quoted second) was in response to v=d/t.
 
  • #35
i'm confused, so in Vf^2 = Vx^2 + Vy^2
and you said that final is 3.60 m/s than what is Vx?
 
  • #36
No. Vx= 3.6 m/s. You need to find Vy before you can find Vfinal.
 
  • #37
oh, than to find Vy = Voy - at?
 
  • #38
Yep.
 
  • #39
than what is the time and Voy
i know that a = -9.8
 
  • #40
hurtingBrain said:
than what is the time and Voy
i know that a = -9.8


Vo is the initial velocity, you have the initial speed [17.0 m/s] and the angle [35.0° above the horizontal] and a hint:


physicsface said:
the sine helps you solve for the vertical component.
 

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