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

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The discussion revolves around solving a projectile motion problem involving a cliff and the application of the quadratic formula. The user attempts to derive the time of flight using the equation Yf = Yinitial + ViT + 1/2 aT^2 but struggles with factoring out T, leading to confusion about possible solutions. They calculate a time of 23.9 seconds but realize they may need to use the quadratic formula instead due to the nature of the equation. Additional inquiries are made about finding more 2D kinematics examples for practice, highlighting a common challenge in mastering these concepts. The conversation emphasizes the importance of correctly applying mathematical methods in physics problems.
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:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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