Yes because the rocket travels in a parabolic shape. The overall time it takes includes both when it rises and falls and since you want the tallest height it flies, you want where time is at half of the overall duration in travel.
Not true at all.
Intelligent people always go for the more efficient method of solving problems.
As for putting x = 10 into the polynomials, I don't understand this step:
Can you please tell me how you derive that?
Maybe I'm getting the wrong idea from this, but you're saying that if you divide ANY polynomial expression through synthetic division, I would always get 10 as one of my roots?
Yes. A proof of the algorithm would better help me understand the workings behind this algorithm.
EDIT: Whats the difference between regular long division using base 10 and synthetic division? Don't they both use a decimal based system?
I'm always curious as to how this works? I mean, it's amazing that, even though dividing by binomials, that this sort of mechanism works but I don't understand quite the qwerks behind this.
Can someone fulfill my curiousity?
@HallsofIvy - So that does mean infact that at 6 seconds, she is closest to the bus..?
Ok so the answer to the problem is 7.0m ..
From what Astronauc just told me, I'm wondering is the problem really this easy?
Well we know from what he said was that it the woman is 6 seconds closest to the...
The question is as I have typed. Nothing was left out.
@Astronuc - So presumably, the bus is closest at 6 seconds then?
@Steve23063 - So after doing that process, if she does indeed catch the bus, does that mean I wouldn't have to find her frustration distance?
By trying your method:
Bus:
Δd...
I encountered yet another difficult one but this time, this problems a bit funny because there's 2 different cases that need to be put into consideration and I don't know exactly how to approach this question.
It goes like this:
"A Pedestrian is running at her maximum speed of 6.0m/s to catch...
Ok, so does this look right?
I foudn the final velocity first because it's shared among stone #1 and stone #2.
2aΔd = Vf^2-Vi^2
2(-10m/s)(120m) = Vf^2 - (70m/2)^2
Vf^2 = 2500
Vf = 50m/s
Using this, we set up the first equation to express the position as a function of time for stone 2:
Δd =...
@Berkeman - If the two accelerations are constant, does that mean the velocities are the same?
And yes, that's the equation I meant. Sorry for the mix-up.
When you say express the position as a function of time, do I equate the two equations afterward to determine the time in which the...
I seem to have some irregularities with a specific problem which goes like this:
"A stone is thrown upward from the edge of a cliff, with a speed of 70m/s. If, when it is 120m above the hand on its return journey, a second stone is dropped over the edge of the cliff, when and where will the...
Well think of it first. Your textbook should at least very basically tell you what the formulas you should use for your homework.
In general, it always follows the equation y = mx + b. From here, you can derive all your basic information for a graph. If you want standard form, use y_2 - y_1 = m...