Need Antiderivative Help!!!

Okay, yeah, I know I am the "math geek" but we've put 3 minds to this problem, and we keep coming up with 2 non-equal constants having to equal one another in order to make the equation "true". If you are NOT math-smart, LEAVE NOW. This problem is lengthy...

Okay, here's the deal: 2 balls are thrown upward from the edge of a cliff 432 feet above the ground. The first is thrown with a speed of 48 f/s and the other is thrown a second later at 24 f/s. Do the balls ever pass each other? Note, other pertinent information is: we are following the work in the book which states: The motion is vertical and we choose the positive direction to be upward. At time T the distance above the ground is s(T) and the velocity v(T) is decreasing. Therefore the accelleration must be negative and we have a(T) = dv/dT = -32 (32 being the gravitational force). Thus the antiderivative v(T) = -32T + C This much should work for both the first and second balls. Then, given that v(0) = 48 {for the first ball} we see that v(T) = -32T +48 [Which becomes v(0) = 24 for the second ball, and v(T) = -32T + 24 for the second ball].

Now, we know the heights have to be equal in order for them to "pass" eachother, but the problems we encounter are: The second ball is thrown ONE SECOND later than the first, and the second antidifferentiation of v(T) gives us the formula for the maximum height reached, not any height over a period of time T. SOMEONE PLEASE HELP!!!

anatomy/genetics major here, sorry.
anatomy stands for: a non arithmetician try other members, Yes!
:grinno:

hate math, but would help if i could.

If you are NOT math-smart, LEAVE NOW. This problem is lengthy...



gooooodbye! :grinno:

:iamwithst:

grrr, I know I can solve this problem...but I do not have the time at work sorry :(
Maybe when I get home

Ball One
a(t) = -32
v(t) = -32 t + 48
s(t) = -16 t^2 + 48 t + 432

Ball Two
a(t) = -32
v(t) = -32t + C
v(1) = -32 + C = 24

C = 56
v(t) = -32 t + 56
s(t) = -16 t^2 + 56 t + C
s(1) = -16 + 56 + C = 432
C = 392
s(t) = -16 t^2 + 56 t + 392

Ball One
s(0) = 432
v(t) = -32 t + 48
v(0) = 48
a(t) = -32

Ball Two
s(1) = -16 + 56 + 392 = 432
v(t) = -32 t + 56
v(1) = 24
a(t) = -32

-16 t^2 + 48 t + 432 = -16 t^2 + 56 t + 392
48 t = 56 t -40
0 = 8t - 40
t = 5

Ball One
s(5) = 272
v(t) = -32 t + 48
v(5) = -112

Ball Two
s(t) = 272
v(t) = -32 t + 56
v(5) = -104

ok, one thing not stated, i am assuming that the mass of the 2 balls are the same, so therefor the force of gravity will effect each ball the same way. I havent quite figured it out, but just looking from the data.... yes, they will pass each other. the first ball, when at its peak, will still have 48 feet between it and the 24 fps ball. now, the 24fps ball is going half the speed, so it would seem to be logical to conclude that it doesnt have enough momentum to carry it the same height as the 48fps ball. ususing the idea, that the 48fps ball travels for 10 seconds, it goes 480 feet. the 24fps ball only goes 240 feet. half the distance. they would meet, if the 48fps ball was launched 5 seconds before the 24fps ball. the 48fps would be at its half mark, 240 feet as the first 24fps ball was reach its peak, and then they would meet. as it stands, the 48fps ball would be at 432 feet when the 24fps ball reaches its peak of 240. now, assuming they fall the same speed they went up... the 48fps ball will catch up to the 24fps ball and pass it on the way down, 24 feet above the ground. at least, thats what popped into my head when I looked at it. I have never been one to do equations, they just seem to pop into my head witht he answer. this may not be correct... but my brain got me a A in trig and calc.

one way of looking at it, the 24fps second ball travels 10 seconds, but the 48fps ball only travels 9... because the second difference in time, has already allowed the 48fps ball to reach its peak, and travel back down, 48 feet. the 48 fps ball will be in the air for 9 seconds while the 24fps will still be for 10. they'll pass :)


hopefully, this post might be easier to understand and prove right, or wrong.

We got the problem solved!!! Yay!!!

This is only for a calc course, so the weights are assumed equal, and velocity is assumed to be infinite. The problem we encountered was in the fact the balls were on "different time schedules" and we were evaluating the change in time at the first antiderivative instead of the second one (the second antiderivative produces the formula for height given an amount of time). Thanks muchly to all who responded!!! :D

ok, so whats the answer, will they meet? and if so, when?

ok, one thing not stated, i am assuming that the mass of the 2 balls are the same, so therefor the force of gravity will effect each ball the same way.

:eek7: It doesn't matter what the mass of the objects are as far as gravity is concerned. 9.8 meters per second squared - all the time. Mass and surface area will make a difference with air resistance but has nothing to do with the effects of gravity.

They will meet about 160ft. below the edge of the cliff after about 5 seconds (this is just what I remember... I forgot to write down the solution to bring home with me).

I'm so fucked once I get to college.

I'm so fucked once I get to college.


:cwm27: :iagree:

didn't want to do all the math out, but here is what i think it is:

ball 1: D=48t+1/2(-9.8)t^2
ball 2: D=24(t-1)+1/2(-9.8)(t-1)^2

set the D's equal

48t+1/2(-9.8)t^2=24(t-1)+1/2(-9.8)(t-1)^2

solve and you get the time they pass each other. And how could u not know the mass of the ball has no bearing on the acceleration?? Isn't that liek the first thing u learn in physics?

I'm so fucked once I get to college.


You've already been college, bro.

Besides, you're taking engineering. All you need to know is that metal melts and how to make me an MP5 :p

I'm so fucked once I get to college.
Oh well, at least you can impress the ladies by being able to truthfully claim that you have a "mad tyte" pad.


Or you could just do Accounting- "1 dollar muahaahahaha, 2 dollar muahahahahahaha, 3!!!!! dollar muahahahahahaha"

Then I'd be you, Jared :eek:

Fine Rhys. Univerrrrrrrrrrrrrrrrrrsity.

Wrong word to be repeated.

Say after me:

Heckleeeeerrr annnddd Kooossshh.