Simple Kinematics Problem - Is the book wrong?

[Nano-Passion]

Simple Kinematics Problem -- Is the book wrong?

Homework Statement

Is the book wrong or do I have faulty logic?

Here is the problem:

An astronaut on a distant planet wants to detrmine its acceleration due to gravity. The astreonaut throws a rock straight up with a velocity of +15m/s and measures a time of 20s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet?

Homework Equations

t=(v-u)/a

The Attempt at a Solution

The solutionn for this problem is 1.5m/s^2. This solution is assumed on the premise that the time going up is the same as the time taken going down. So that time during ascent is 1/2t. Therefore it is 10s.

But that strikes me as rather unintuitive. If you look at the equation t=(v-u)/a, the time taken during ascent would be different than time taken during descent because the rock is thrown with a initial velocity (u) of 15m/s

 

[iRaid]

V_f = V_i + at

So we can make V_f=0 if we make the time half (when the velocity = 0 that will be at the top)

Therefore: t=20/2 = 10

-V_i = at

Answer:?

 

[Nano-Passion]

V_f = V_i + at

So we can make V_f=0 if we make the time half (when the velocity = 0 that will be at the top)

Therefore: t=20/2 = 10

-V_i = at

Answer:?

The answer is 1.5m/s^2

But how could you just make the time half when the time the rock ascents should be different then the time it descents according to t=(v-u)/a

 

[iRaid]

Is that the answer your book gives you?

 

[iRaid]

http://i51.tinypic.com/2u8jomr.png
I made a picture hope that helps you.

 

[tedbradly]

Let us say a rock is thrown up with velocity v_o. Further, assume the only force acting on it is gravitational.

For the first half (going up) we have:
v_1(t) = v_0 + at
When the rock is at its maximum height, it has zero velocity. Thus, t_1, the time spent going up, equals:
0 = v_0 + at_1 \rightarrow t_1 = -\frac{v_0}{a}

During this trip, the ball traveled a distance d - x_0:
x_1(t) = x_0 + v_0 t + \frac{1}{2} a t^2
x_1(t_1) = d = x_0 + v_0 t_1 + \frac{1}{2} a t_1^2

We then have a second problem to set up while it goes down. x'_0 will be this problem's initial displacement, equal to d. The total time it takes to go down will be when x_2(t) = x_0 (when it returns to the point it was thrown). And of course, v'_0 = 0.

x_2(t) = x'_0 + v'_0 t + \frac{1}{2} a t^2
x_2(t) = d + \frac{1}{2} a t^2
x_2(t_2) = x_0 = d + \frac{1}{2} a t_2^2 \rightarrow t_2 = \sqrt{\frac{2(x_0 - d)}{a}}

We then plug in the equation for d, which is in terms of quantities from the first half of the problem:
t_2 = \sqrt{\frac{2(x_0 - d)}{a}}= \sqrt{\frac{2(x_0 - [x_0 + v_0 t_1 + \frac{1}{2} a t_1^2])}{a}}= \sqrt{\frac{-2v_0 t_1 - a t_1^2}{a}}

Almost done. So we then distribute that a onto each part:
t_2 = \sqrt{-2t_1\frac{v_0}{a} - t_1^2}

But look all the way at the top! - v_0/a is equal to t_1! Thus:
t_2 = \sqrt{2t_1 t_1 - t_1^2}= \sqrt{2t_1^2 - t_1^2} = \sqrt{t_1^2} = t_1
t_2 = t_1

 

[Nano-Passion]

Is that the answer your book gives you?

Yes.

http://i51.tinypic.com/2u8jomr.png
I made a picture hope that helps you.

I know time is split in half, but I my logic tells me that the time taken in ascent would be different from the one in descent because there is initial velocity upward against acceleration involved.

 

[SteamKing]

But if you look at the problem this way, at the point of the highest ascent of the rock, the velocity is zero. When the rock falls back toward the asteroid and is passing the astronaut's had, the acceleration on the rock due to the gravitation of the asteroid has produced a certain velocity in the rock. How fast is the rock traveling when it passes the hand of the astronaut? It must be the same as when the astronaut initially threw the rock upward. Hence, the time of travel upward for the rock is the same as the time of travel downward, which is what the calculation of tedbradly shows. The rock was thrown up with an initial velocity and is acted on solely by the gravity of the asteroid while aloft.