Thread: Interesting gravity problem.

1. Interesting gravity problem.

Suppose the acceleration of gravity of planet X is one-half that on the earth. Determine the initial velocity of the ball on the earth (in terms of $v_0$) so that the maximum height attained is the same as on planet X.

What I tought was, first to find what the maximum height in planet X is.
So, having g as my constant for gravity I have:

On Earth : $a(t)=-g$ ; and on Planet X: $a(t)=-g/2$
Then when setting the velocity on Planet X equal to zero, and plugging that $t$ in the position function : $s(t)=(-1/4)gt^2 + v_0t$; supposing that the initial position is zero.

I get: $[(v_0)^2]/g$ as the maximum height on planet X.

Once I get that answer I try to set it equal to the position function of Earth, but it doesn't seem to work out well. And I'm not even sure if that's what I should do.
Any help would be greatly appreciated.

2. Originally Posted by Arturo_026
Suppose the acceleration of gravity of planet X is one-half that on the earth. Determine the initial velocity of the ball on the earth (in terms of $v_0$) so that the maximum height attained is the same as on planet X.

What I tought was, first to find what the maximum height in planet X is.
So, having g as my constant for gravity I have:

On Earth : $a(t)=-g$ ; and on Planet X: $a(t)=-g/2$
Then when setting the velocity on Planet X equal to zero, and plugging that $t$ in the position function : $s(t)=(-1/4)gt^2 + v_0t$; supposing that the initial position is zero.

I get: $[(v_0)^2]/g$ as the maximum height on planet X.

Once I get that answer I try to set it equal to the position function of Earth, but it doesn't seem to work out well. And I'm not even sure if that's what I should do.
Any help would be greatly appreciated.
If $v_{E0}$ is the initial velocity on earth, then the position of the ball on earth is given by $s(t)=v_{E0}t-\tfrac{1}{2}gt^2$.

The maximum height is reached when $t=\frac{v_{E0}}{g}$ (verify).

Thus, the maximum height is $\frac{v_{E0}^2}{2g}$ (verify).

If the maximum heights on earth and planet X are to be the same, then $\frac{v_{E0}^2}{2g}=\frac{v_0^2}{g}$. Now solve for $v_{E0}$.

Can you take it from here?

3. That is actually what I considered, but it didn't seem to make sense at first. Now I see. Thanks.

P.S. Your animations are hilarious.