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

In summary, The 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. It lands 23.9 seconds from the foot of the cliff.
  • #1
dolpho
66
0

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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  • #2
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)
 
  • #3
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
 
  • #4
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:
 
  • #5
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.
 
  • #6
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:
 

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

1. What is a 2D kinematics problem?

A 2D kinematics problem is a type of physics problem that involves motion in two dimensions, typically represented by the x and y axes. It involves using equations and principles of kinematics to analyze the motion of an object in terms of its position, velocity, and acceleration.

2. What are some common equations used in 2D kinematics problems?

Some common equations used in 2D kinematics problems include the equations of motion (d = v0t + 1/2at2, v = v0 + at, and v2 = v02 + 2ad), the Pythagorean theorem (a2 + b2 = c2), and vector addition equations (a + b = c and a2 + b2 = c2).

3. How do you approach solving a 2D kinematics problem?

The first step in solving a 2D kinematics problem is to identify the variables given and the variable to be solved for. Then, use the appropriate equation(s) to set up a system of equations and solve for the unknown variable. It is also important to draw a diagram and label all relevant quantities to help visualize the problem.

4. What are some common misconceptions about 2D kinematics?

One common misconception is that acceleration is always constant in 2D kinematics problems. In reality, acceleration can vary in both magnitude and direction. Another misconception is that the x and y components of velocity are independent of each other. In fact, they are related through the Pythagorean theorem and vector addition equations.

5. How can I check my answer for a 2D kinematics problem?

You can check your answer by plugging your values back into the original equations and making sure they satisfy the given conditions. It is also helpful to check that your answer has the correct units and is reasonable based on the given scenario. Additionally, you can use a graphing calculator or online tool to plot the motion and compare it to the given information.

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