>4.0 GPA
>never failed an exam in my life
>just totally bombed my physics exam...
Is this problem unsolvable? I'm pretty sure it's impossible to solve. Not looking for an answer, just want to know if this stupid fucking problem is even POSSIBLE to solve.
"A person standing on a building throws a ball directly upwards. After 5 seconds, the ball passes a window that is 30 meters below the point the ball was thrown. What is the initial velocity of the ball when it was thrown?
You are retarded.
If it's so fucking easy then how do you solve it? Everytime I see this problem asked either the initial velocity is given or the final velocity is given. I just feel like you don't have enough knowledge to solve.
The answer is a formula
>faggot has no deductive skills nor course friends to ask
the aceletarin is known if this is in earth (ovbiuosly is) so get fucked
The answer is 18.5 m/s
This is the easiest of physics equations. You have distance, acceleration, and time. EZ
I can't wait until you get to F=ma
The ball is thrown upwards and reaches a certain height, Y, above the point from which it was thrown. Y=0.5 A t^2 Call that t, t-sub-1.
if then drops (Y+30) = 0.5 A t^2. Call this t, t-sub-2.
t-sub-1 plus t-sub-2 is five seconds.
Solve for Y.
V=sqrt(2*A*Y)
Assumptions: A is Earth-gravity. No air resistance.
Let d be the distance the ball travels upwards, and t the time it takes until the ball reaches the highest point
Then the ball moves (obviously) d meters upwards in t seconds, and then d+30m downwards in 5-t seconds
2 formulas for 2 variables
Not enough data for meaningful answer
Now if the ball had to pass THROUGH the window, THEN you'd have a problem.
For some reason I thought the acceleration when it's thrown up (due to the hand pushing it forward) and acceleration when it falls are different. I'm retarded and forgot that acceleration is same throughout. Damn I'm an idiot.
>bragging about high school gpa
I wasn't bragging, I just wanted to emphasize that I've never bombed an exam before, I realize I screwed this exam up now though
x is the vertical position of the ball. The height of the person is set as 0.
There is a constant acceleration to the bottom of 9.81m/s^2.
v = v0 − t*9.81m/s^2
x = t*v0 − (1/2)*t^2*9.81m/s^2 = t*v0 − t^2*4.91m/s^2
x = −30m
t = 5s
−30m = 5s*v0 − (5s)^2*4.91m/s^2 = 5s*v0 − 122.75m |+ 122.75m
122.75m - 30m = 92.75m = 5s*v0 |/5s
92.75m/5s = 18.55m/s = v0
v0 = 18.55m/s
What the hell is this shit? The equation is simple
D= v0*t +1/2(a)(t^2)
D=-30 (30 meters below the frame of reference)
a = -9.8
t= 5
This is literally middle school algebra except you are given an equation
>exam based education system
This is why I smoke the autistic 4.0 gpa "memorization is learning" morons every test.
Addition: It is not solvable if the gravitational acceleration is not given. Maybe this was the problematic point. I think one can assume that it does happen on the earth.
mate the ball is thrown up. You need twoo equations. Its still easy but you just gpt it wrong.
You don't need two equations. Stop being a brainlet. That's why physics is so cool: the equations are absolute. You most certain can have a D of -30, and you certainly can start with a positive velocity and negative acceleration. All of it lumps into one equation.
Hey if you are finding high school physics difficult then consider pursuing to a non-stem major, I hear journalism is pretty good. Also you may want to stop browsing this board, and come back when you are over 18.
Cheers!
You get the same exact answer as the other poster said: v0 = 18.55m/s
All of you are brainlets except
Doesn't matter if the ball is thrown up. And I don't have it wrong. This is basic physics.
hes right. there are 2 situations here. going up then down. but you use the same reference frame throughout
a_up is the opposite of a_down. you cannot use the same value for both.
if you try to cheat that, youll run into problems later on in physics
"acceleration due to gravity is different going up than it is going down"
Dude I seriously hope you're trolling.
"acceleration is the same in all directions. also acceleration is not a vector."
whos the one trolling?
a_up=-a_down
Since you are either trolling or 100% braindead, I'll put this here for others who don't know:
Acceleration due to gravity is a vector. It is always present, and its magnitude is always 9.81 unless you start to get significantly farther away (orbit) or closer to the center of the earth. Since our frame of reference is (close to)the surface of the earth, this vector is always pointing down, which is in the negative direction. Since these statements are true, acceleration due to gravity always -9.81 m/s and will never change in any basic physics class.
That being said, acceleration due to some other force is completely different. In the OPs scenario, we aren't concerned with the force applied by the person since that becomes 0 as soon as it leaves his hand.
The guy I'm replying to is retarded and should be ignored.
Did I do it right? I'm a brainlet engineer student desu
Yes, if that slop down at the bottom says 18.5 m/s
>Brags about high school GPA
>Brags that he hasn't failed failed an exam in his first semester at big boy school
>Fails his first physics exam
Life's about to get a lot harder for you
Effective gravity on the Earth's surface varies by around 0.7%, from 9.7639 m/s2 on the Nevado Huascarán mountain in Peru to 9.8337 m/s2 at the surface of the Arctic Ocean. In large cities, it ranges from 9.7760 in Kuala Lumpur, Mexico City, and Singapore to 9.825 in Oslo and Helsinki.
I know how to solve These problems but they still bother me every time I see one. If the ball is thrown directly upwards as required by one dimensions kinematics, why doesn't it land on the building again?
Well maybe the guys hand is reaching outward over the ground below
> What is the initial velocity of the ball when it was thrown?
Is he a big guy? One hand throw or two? Under arm or over? Has he done this before? Soy?
Youre combining two equations you fool.
Funny how you all suck so hard at Newtonian mechanics yet are all somehow experts on general relativity
You are correct, i apologie.
that's ameritard education for you people
do you ameritards learn this in college?
because that's 1st year of highschool in EU
This problem is really simple. I saw the picture and assumed it was a trajectory requiring at least parametric equations or something. How do you not know the general formula for displacement? This may not be your fault, it could be the teacher is doing things wrong.
[math] d = s_0 + v_i t + \dfrac{1}{2} a t^2 [/math]
plug and chug from there. If your professor hasn't placed good emphasis on this, then you are going to have to find these equations for every topic you guys cover. It is going to be in the text book.
>arguing over semantics for a question straight out of an introductory course in mechanics
physics classes there is the assumption that the acceleration is always 9.8m/s squared. This can never be an excuse. OP just can't into basic physics which is the weed out for STEM majors apparently.
s should be -30m right?
If you wrote something like this on a physics test I would assure you you would actually get marked for the correct answer but the professor would write a side note for you either being a smart ass or having no life for knowing this much info as an undergrad.
Depends what sign "convention" you use. The gravitational acceleration points downwards and I used a positive value for it (+9.81), therefore positive velocities and distances also point downwards.
I would always make displacement absolute values.
But it isn't!
shh not yet user
for projectile yes. if it's something like a plane or rocket ship that propels itself then no.
...
what if they's on the moon
you divid 32.174 by about 4
Quadratic formula; solve for 'B' using only the positive side with known 'A' , 'C' and 'X'
0. u throw it while it's still
Simply solve [eqn]
\left\{
\begin{align}
y''(t) &= -g \\
y'(0) &= v_0 \\
y(5) &= -30
\end{align}
\right.
[/eqn] for [math]y[/math] and then solve [math]y(0) = 0[/math] for [math]v_0[/math]
Answer is [math]\frac{1}{2} (5 g-12)[/math]
Listen up brainlets:
Distance the ball will travel upwards:
[eqn]H_{max} = \frac{v_{0}^2}{g}[/eqn]
Time the ball will travel upwards:
[eqn]t_{a}=\frac{v_{0}}{g}[/eqn]
From:
[eqn]S = v_{0}t_+\frac{gt^2}{2}[/eqn]
we get final equation (ball moving downwards):
[eqn]H_{max} + 30 = \frac{g(5 - t_{a})^2}{2}[/eqn]
[eqn]\frac{v_{0}^2}{g} + 30 = \frac{g(5 - \frac{v_{0}}{g})^2}{2}[/eqn]
result is ~16 m/s
HAHAHAHAHAHA
Holy fucking shit I can't believe most of you are in university.
>answer is solved in the 5th comment
>55 replies
This is why this board is trash
no, this is why this board is trash
if you don’t use calculus and instead use a memorized equation to solve this, i am loling at your life
object is always under the same acceleration dipshit
acceleration due to gravity is always pointing down
yes and posts like yours are making it better
thank you
user just ask your teacher. You can't be in that big of a high school.
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