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/sci/ - Science & Math

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5059346 No.5059346 [Reply] [Original]

I guess everyone here is a genius, but I really need help as I just started kinematics. I can do basic problems except I don't understand the right time to use the equations, i can't use them as tools to my benefits,for example If i throw a ball up in the air and it goes 80km/h how high can i throw it? Please help i would really appreciate it,

>> No.5059347

Earth's gravity = 9.8m/s²

>> No.5059353

>>5059347
Thanks, but I know that, i can know the measurements i just don't have the logic to know where to put the vectors and scalars i have

>> No.5059356

>>5059346
Take that Physics book, look at some problems and start putting those kinematics into practice. You will learn no other way.

>> No.5059376

>>5059356
is it kind of a trial and error thing until i discover patterns and when i look at a problem i can go "oh that didnt work last time, but i know what did"?

>> No.5059381
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5059381

It really helps if you've taken some basic calculus and understand how all the kinematic formulas (like <span class="math">x(t) = x_0 + v_0 t + \frax{1}{2}at^2[/spoiler]) are derived.

>> No.5059385

>>5059381
damn i can't type, i meant <span class="math">x(t) = x_0 + v_0 t + \frac{1}{2}at^2[/spoiler]

>> No.5059397

>>5059376
Yes.

>> No.5059406

>implying everyone on /sci/ is a genius

Take it from someone who lurks this board everyday: not everyone here is a genius. Although they are, by far, the most intelligent guys on 4chan.

>> No.5059408

>>5059385
Yea im good with math so i derived all the equations i learned, all i hav till now is

a=delta(v)/delta(t)
v= v(i)+at
d=t[(v(i)+v(f))/2]
d=v(i)t+1/2at^2
and
v(2)^2=v(1)^2 + 2ad

>> No.5059414

>>5059406
Who are you talking to?

>> No.5059434

>>5059414
OP

>> No.5059453

you can always graphically solve a problem

>> No.5059455

>>5059346
The easiest way to do kinematics is to:
1) Write down every variable you know (acceleration, v_0, etc.)
2)Write down what variable you are looking for
3)Then look at the four kinematic equations you use:

d_f = a/2*t^2 + v_0*t + d_0
v_f = a*t + v_0
v_f^2 = v_0^2 + 2*a*d
d = 1/2(v_0 + v_f)*t

>> No.5059463

>>5059455
and then what? like one question was missing 2 variables and i found them both with same equation?

>> No.5059469

Alright, OP, let's take a look at the stuff we know. We have the following information:

<div class="math">v_0 = 80 km/h = 22.22 m/s</div>
<div class="math">v = 0 m/s<div class="math"> (as you are interested in when its going to stop. Note that there will be a moment where the ball's speed is 0 before it starts a positive acc.
a = -9.81 m/s^2

I assume that you know the basic formulas. Take a look at this one:

v^2-v_0^2 = 2as

where s is distance. How would you use this formula to solve your problem?</div></div>

>> No.5059471

>>5059469
I tried using latex for the first time. I failed beautifully.

>> No.5059474

>>5059463
Oops didn't type last step
4)Find the equation that has all of the variables you know and only one that you don't know. You just solve the problems variable by variable until you answer everything the question asks for.

>> No.5059489

You know you guys are overcomplicating this problem by far, right? Let me try to explain without using latex.

v_0 = 22.22 m/s
v = 0 m/s
a = -9.81 (m/s)^2

v^2 - v_0^2 = 2as

22.22^2 - 0^2 = (2 * -9.81)s

493.77 = -19.62s

s = -25,167 m

And since a negative answer doesn't make any fucking sense, just make it positive, be finished with your homework, and masturbate.

>> No.5059504

>>5059347
which depends on where you are on Earth
it's relative
but that number is the most widely accepted standard

>> No.5059533

OMFG guys thank you, I did the question: how high can you throw a ball if you throw it 80km/h upwards, but i still don't understand why u have to substitute -9.8m/s^2 instead of positive, and why the final velocity is 0

>> No.5059538

>>5059474
Thanks a bunch, i just wanted to hear that now i know how to use the equations

>> No.5059548

>>5059533
Because the acceleration is negative. The ball is not gaining speed, it is losing speed.

Furthermore, the final velocity is zero because that is the point you are interested in. The final velocity, the highest point of your throw, will be zero. Afterwards, it will start gaining speed and fall towards the ground.

>> No.5059589

>>5059548
So final velocity in similar scenarios will always be at the highest point of the body's travel? and also i understand the difference between distance and displacement but how come my teacher said the arrow isnèt necessary when writing the equation? theres a difference between the building and the floor and the total path length can someone help me understand? thanks

>> No.5059608

>>5059589
Yeah, all similar scenarios will have the highest point of an object's path when thrown directly up be at 0 m/s. That's the point where the object is in a little bit of hang time and gravity begins pulling it back to Earth.

>> No.5059617

>>5059504
shut the fuck up

>> No.5059621

>>5059608
>pulling
Pushing*

>> No.5059634

>>5059617
s-sorry

>> No.5059642

Ok now, i got a question saying a person speeds past a gas station at 190km per h so around 52meters per second. The cops see her and start to chase her and the cop car has acceleration of 4.9meters per second squared, how far will they go in order to catch up to the girl? i got 285 meters solving the equation like 3 different ways with a system as well, and the answers say it was 1.1km. If anyone is free please tell me whos wrong me or answers

>> No.5059662

For rectilinear motion of particles, that is a point in space moving straight, generally the motion is described by

x=x(t), v=dx/dt , a=dx^2/d^2t = v dv/dx

all your formula can be derived from these

>> No.5059677

>>5059617
Physics isn't just about the numbers. Acceleration due to gravity isn't a constant. Even though, OP won't deal with problems where he needs to understand that yet, it's good that OP learns it anyways. There is nothing wrong with knowledge or accuracy.

>> No.5059683

>>5059642

v*t=a*t^2/2, t=2v/a, s=v*t

v=52.8 m/s, a=4.9 m/s^2

s=1137 m

>> No.5059688

>except I don't understand the right time to use the equations

you are given N values
Find the equation that has N-1 variables that match with the given values
Solve

>> No.5059697

>>5059642
You are dealing with the motion of two cars. One is traveling at constant velocity, the other starts from rest and accelerates to a velocity that will be greater than the first cars constant velocity. The solution in the book is correct.

>>5059683
Doing the problem for him won't teach him how to do the problem.

>> No.5059835

>>5059683
how did u get those equations, and yea thats the right answer

>> No.5060032

>>5059835

s=v*t describes the distance travelled at velocity v after time t
s=a*t^2/2 describes the distance travelled at acceleration a after time t

both travel the same distance s because they meet after time t
so you set the two equations equal and solve for t and you get t=2v/a
now you calculate s by plugging the t you found into either s=v*t or s=a*t^2/2
the obvious choice is s=v*t because it is easier to calculate

that's really all there is to it