Simple 2-D Projectile Motion Question

[Austin Chang]

Homework Statement

A ball is thrown horizontally from the top of the cliff of height H. It hits the ground a distance H from the bottom of the cliff. What was the initial speed?

X coordinate
v₀ = ?
a = 0 m/s²
x₀ = 0
x_f = H
Y coordinate
v₀
a = -9.8 m/s²
y₀ = H
y_f = 0

Homework Equations

a(t) = a
v(t) = v₀ + at
s(t) = x₀ + v₀t + 1/2*at²

The Attempt at a Solution

H = 0 + v₀t
0 = H + 0 + 1/2 * (-9.8) t²
-H = 1/2*(-9.8)
H/4.9 = t²
(H/4.9)^1/2 = t

 

[haruspex]

Homework Statement

A ball is thrown horizontally from the top of the cliff of height H. It hits the ground a distance H from the bottom of the cliff. What was the initial speed?

X coordinate
v₀ = ?
a = 0 m/s²
x₀ = 0
x_f = H
Y coordinate
v₀
a = -9.8 m/s²
y₀ = H
y_f = 0

Homework Equations

a(t) = a
v(t) = v₀ + at
s(t) = x₀ + v₀t + 1/2*at²

The Attempt at a Solution

H = 0 + v₀t
0 = H + 0 + 1/2 * (-9.8) t²
-H = 1/2*(-9.8)
H/4.9 = t²
(H/4.9)^1/2 = t

Good so far. Why did you stop? You have two equations, each involving t. Do you expect to have t in the answer?

 

[hmmm27]

Looks pretty confusing to me, but I'm new to this. You don't mind if i ask a few questions...

Homework Equations

a(t) = a
v(t) = v₀ + at
s(t) = x₀ + v₀t + 1/2*at²

I don't see how the first and second equations are "relevant". (Actually I'm not entirely sure what the first equation is supposed to mean)

The third equation looks complicated... should that be $d_t = d_{init} + v_{init}t + \frac {at^2}{2}$ ?

The Attempt at a Solution

H = 0 + v0t
0 = H + 0 + 1/2 * (-9.8) t2
-H = 1/2*(-9.8)
H/4.9 = t2
(H/4.9)1/2 = t

Yeah well, my first attempt ended up with the answer as $v=4.9t$

 

[Austin Chang]

Good so far. Why did you stop? You have two equations, each involving t. Do you expect to have t in the answer?

Do i get H in my answer because I thought we were suppose to get a quantitative number

 

[Austin Chang]

Looks pretty confusing to me, but I'm new to this. You don't mind if i ask a few questions...I don't see how the first and second equations are "relevant". (Actually I'm not entirely sure what the first equation is supposed to mean)

The third equation looks complicated... should that be $d_t = d_{init} + v_{init}t + \frac {at^2}{2}$ ?
Yeah well, my first attempt ended up with the answer as $v=4.9t$

I just typed everything down on my page LOL, but answering your question the first and second question are me just integrating upwards because I don't like memorizing

 

[Austin Chang]

Do i get H in my answer because I thought we were suppose to get a quantitative number

and I got (4.9H)^1/2 = v₀

 

[PeroK]

Do i get H in my answer because I thought we were suppose to get a quantitative number

You get H in your answer if the answer depends on H.

Can you see a reason why it must depend on H in this case?

If not, then trust your maths.

 

[hmmm27]

and I got (4.9H)^1/2 = v₀

which is what I got the second time I juggled the equations around.

 

[Austin Chang]

You get H in your answer if the answer depends on H.

Can you see a reason why it must depend on H in this case?

If not, then trust your maths.

No because a question a while back asked if a ball is dropped from a certain height h and a ball thrown at a certain velocity hits each other in the middle with the same velocity. Find the height of h.
So I am confused because some questions you don't need height and others you do. Is this similar to the one variable one equation sort of thing

 

[hmmm27]

It's up to the course you're taking and the teacher, what they want.

My take would be that the answer should be a speed (because it's explicitly stated, so duh) written as a function of H (because it already exists in the question, so using 't' would just be superfluous)

 

[PeroK]

No because a question a while back asked if a ball is dropped from a certain height h and a ball thrown at a certain velocity hits each other in the middle with the same velocity. Find the height of h.
So I am confused because some questions you don't need height and others you do. Is this similar to the one variable one equation sort of thing

Well, you just have to trust your maths. If you get a value for initial speed that depends on H, then that must be right.

In this case, it must depend on H.

 

[hmmm27]

4.9t is also a "correct" answer.

His question is in what form should the answer be in, not what does random linear algebra produce.

 

[haruspex]

Do i get H in my answer because I thought we were suppose to get a quantitative number

Dimensional analysis shows H must feature in the answer. You are only given H and g, a distance and an acceleration, and you are asked to find a velocity. You cannot find a velocity only given an acceleration since there is no way to disentangle the distance element of acceleration from the time-squared element. You would only be able to find other accelerations. But given a distance H and an acceleration g you can derive velocities of the form √(Hg).

 

[haruspex]

4.9t is also a "correct" answer.
.

no it isn't. H was given in the question statement, so can validly be referenced in the answer. t was an unknown invented as part of the solution, so cannot appear in the answer. If that were allowed, you might just as well invent the unknown v for initial velocity and quote that as the answer!

 

[hmmm27]

I often use quoties to indicate sarcasm or non-truth, among other things; in this case a "technical correctness".

H was given in the question statement, so can validly be referenced in the answer. t was an unknown invented as part of the solution, so cannot appear in the answer. If that were allowed, you might just as well invent the unknown v for initial velocity and quote that as the answer!

Right, so what you said in your previous post, and what I said a few posts back, immediately after your post where you seemed (to me) to be promoting ambiguity in the answer. both of which earned us 'likes' from the OP who has gone away happy.

 

[haruspex]

I often use quoties to indicate sarcasm or non-truth; in this case a "technical correctness".

It is no more correct, even technically, than answering v₀. Anyway, please try to avoid posts which are likely to confuse students.

Right, so what I said a few posts back

True, but not what you first posted.

 

[Austin Chang]

It is no more correct, even technically, than answering v₀. Anyway, please try to avoid posts which are likely to confuse students.

True, but not what you first posted.

Thanks guys both of you were really helpful. :P

 

[hmmm27]

It is no more correct, even technically, than answering v₀.

Betcha some of his class returned $4.9t$ as an answer, since that's the easiest algebraic route to take. And bet you they get part marks for it, too. So yes, it would be more correct than $v_0$

And I bet some people might think your "dimensional analysis" remark refers to nothing more meta than Cartesian coordinates.

Anyway, please try to avoid posts which are likely to confuse students.

Turned out the student had it right the first time but was confused by other solutions to similar problems. I was confused by his using the Relevant Equations section as a scratch pad, instead of cutting/pasting vanilla formulae from the textbook (which, through application of dimensional analysis I figger is what should go in there).

True, but not what you first posted.

You Quoted a fresh post while I was editing it. I did actually hang around awhile, waiting for you to commit your comment. My bad for not getting it right the first time, but I don't recall there being any major differences pre/post-edit.

 

[haruspex]

a(t) = a

Actually I'm not entirely sure what the first equation is supposed to mean)

I read it as saying that acceleration, in principle a function of time, is here a constant.

And bet you they get part marks for it, too.

Not from me they wouldn't.

I bet some people might think your "dimensional analysis" remark refers to nothing more meta than Cartesian coordinates.

Eh?

 

[hmmm27]

I read it as saying that acceleration, in principle a function of time, is here a constant.

Well, that's how I read it too, but it didn't make sense in context of Relevant, since it's already a given from the "top of a mountain" parameter, and none of the other formulas allow for $a$ as a non-constant.

Not from me they wouldn't.

My kneejerk reaction is that a course that would require the student to map everything onto explicit x,y coordinates - when there's a perfectly reasonable simpler framework description("horizontal", "vertical") given that makes for a clearer solution and answer, gives part marks. Not an insult, just saying.

Eh?

Okay, so what do you mean by "dimensional analysis" ?

 

[haruspex]

so what do you mean by "dimensional analysis" ?

http://theory.uwinnipeg.ca/physics/intro/node5.html
... but there is an unfortunate typo at the end. The last two t² should be T²:
$\frac 12 [a][t^2]=\frac L{T^2}T^2=L$

 

[hmmm27]

oh, okay, so that's solving for dimensions as well as the numbers, as a separate process rather than inline. Fair enough : I thought you meant something more metaphysical in "problem-solving space"(for want of a known proper term). (As far as mistakes in the linked text goes, L, while obvious in context, is missing a defining line)

So how would that be applied to the thread's problem ?

Easy enough to write $d = vt$ and $d = at^2/2$ as $L=LT/T$ and $L=(L/T^2)T^2$ for the equations bit, but they're standard equations that come known prebalanced...

So in the parameters there's two $L$'s (that are equal), an implied $L/T^2$ and a missing $L/T$ ? and that translates into you knowing that the output will be in the form ...

 

[haruspex]

So how would that be applied to the thread's problem ?

[H]=L
[g]=LT^-2
[velocity to be found]=LT^-1
There is no way to manipulate the LT^-2 alone to obtain a LT^-1. Allowing use of the L (i.e. [H]) we can write (LT^-2L)^1/2.

 

[hmmm27]

... and you found the dimensions of the answer without bothering with differential equations at all.

<adds "Dimensional Analysis" to the todo list>