Thread: Gravity on Jupiter

1. Gravity on Jupiter

If a rock falls from a height of 20 meters on planet Jupiter, its height H (in meters) after x seconds is approximately:

H(x) = 20 - 13x^2

(a) When is the height of the rock 15 meters?

(b) When does the rock strike the ground?

NOTE: The question comes from my precalculus book. Chapter title: FUNCTIONS.

2. Originally Posted by symmetry
If a rock falls from a height of 20 meters on planet Jupiter, its height H (in meters) after x seconds is approximately:

H(x) = 20 - 13x^2

(a) When is the height of the rock 15 meters?

(b) When does the rock strike the ground?

NOTE: The question comes from my precalculus book. Chapter title: FUNCTIONS.
You should be able to do these.

a) $15 = 20 - 13x^2$

$-5 = -13x^2$

$x^2 = \frac{5}{13}$

$x = \sqrt{\frac{5}{13}}$

b) Same thing only now H(x) = 0. I get $x = \sqrt{\frac{20}{13}}$

-Dan

3. ok

I totally get it.

So, I needed to REPlACE H(x) with 15 meters for a and 0 for part b to simplify.

Thanks!

4. Originally Posted by symmetry
If a rock falls from a height of 20 meters on planet Jupiter, its height H (in meters) after x seconds is approximately:

H(x) = 20 - 13x^2

(a) When is the height of the rock 15 meters?

(b) When does the rock strike the ground?
It never strikes the ground, Jupiter does not have a surface (as we understand it,
and certainly has nothing resembling one any where near where the acceleration
due to gravity is close to 26 m/s^2).

RonL

5. Originally Posted by CaptainBlack
It never strikes the ground, Jupiter does not have a surface (as we understand it,
and certainly has nothing resembling one any where near where the acceleration
due to gravity is close to 26 m/s^2).

RonL
Actually there is some indication that there may either be a diamond core (which seems a bit far-fetched, but possible) or a solid metallic hydrogen core (which I personally think is more likely.) But agreed, we can obviously only speculate at this point.

-Dan