How Can I Solve 2D Kinematics Problems Using the Quadratic Formula?

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Homework Help Overview

The discussion revolves around a 2D kinematics problem involving a projectile launched from a cliff. The original poster presents a scenario where a projectile is shot at an angle, and they are attempting to determine the distance it lands from the base of the cliff. The problem includes elements of projectile motion and the application of kinematic equations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use the kinematic equation to solve for time but expresses confusion about the validity of their approach after factoring out a variable. They question whether they lose possible solutions by doing so and consider using the quadratic formula instead.
  • Some participants question the steps taken by the original poster, particularly regarding the treatment of time as a variable and the implications of setting it to zero.
  • Others suggest reconsidering the setup of the equation and the interpretation of the results, particularly in relation to the quadratic formula.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the methods to solve it. There is no explicit consensus on the best approach yet, but there is a shared understanding of the challenges involved in applying the quadratic formula in this context.

Contextual Notes

Participants note the difficulty in finding additional examples of 2D kinematics problems for practice, which may be contributing to the confusion. There is also mention of differing methods learned in calculus that may complicate the understanding of this kinematics problem.

dolpho
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Homework Statement



A projectile is shot from the edge of a vertical cliff 60.0meters above the ocean. It has a speed of 100m/s and is fired at an angle of 35 degrees above horizontal. How far from the foot of the vertical cliff does the projectile land.

Homework Equations



Yf= Yinitial + ViT + 1/2 aT^2

The Attempt at a Solution



Ok so I can solve this problem but what really confuses me is why I can't solve for T using Y= Yinitial + ViT + 1/2 aT^2. I think they give us all the variables to solve it this way.

Yf = 0, Y initial = 60, Vi = 57.4, a = -9.8m/s^2

So I set it up like, 0=60+57.4T-4.9t^2
From there I factor out a T, -60=t(57.4- 4.9t)
t= -117, -/ 4.9= t
=23.9 seconds.

(I assume we lose possible solutions if I factor out the T, so instead I have to use the quadratic formula? Not sure though)

And just a random question, Does anyone know where they gave a bunch of 2d kinematic examples and ones regarding force? I've been looking all day but can't really find a bunch of examples for practice. Thanks guys!
 
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hi dolpho! :smile:

(try using the X2 and X2 buttons just above the Reply box)
dolpho said:

Homework Statement

So I set it up like, 0=60+57.4T-4.9t^2
From there I factor out a T, -60=t(57.4- 4.9t)
t= -117, -/ 4.9= t
=23.9 seconds.

sorry, not following that :confused:

use the usual quadratic equation formula (the negative result is the time t at which it would have had to be thrown from the foot of the cliff to reach the top at 35° at t = 0)
 
Oh woops, sorry!

-60 = t (57.4 - 4.9t)
one of the t's goes away t=0
-117.4 = -4.9t

Which is like 23 something which isn't the right one! :| Argggg guess I'll have to use the quadratic formula...I hate that thing
 
dolpho said:
-60 = t (57.4 - 4.9t)
one of the t's goes away t=0
-117.4 = -4.9t

no, you can't do that, t = 0 isn't a solution, is it? :wink:
 
tiny-tim said:
no, you can't do that, t = 0 isn't a solution, is it? :wink:

No :( But it works sometimes! when I have like... 0 = 4t + 12t2 and I factor out a t.

Then it's t=0, and -4 = 12t

-4/12 = -.33 and that comes out to be the right answer on the homework or something. It's hard to find an example but it's messing with my mind! I'll try to look for one. :) Ty for help :D

My calculus class is solving acceleration and velocity problems by taking derivatives and factoring. I think that's why I'm so confused.
 
dolpho said:
No :( But it works sometimes! when I have like... 0 = 4t + 12t2 and I factor out a t.

Then it's t=0, and -4 = 12t

yes, you certainly can factor out t if the LHS is 0, because 0/t is still 0 :smile:

but if the LHS is 60, you get 60/t = 57.4 - 4.9t :cry:​
 

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