- #1

- 1

- 0

- Thread starter annyaroo
- Start date

- #1

- 1

- 0

- #2

- 107

- 0

There are two ways to do this probelm:

You need to set up two equations:

First, the kangaroo's velocity is zero at the top of the jump so you know that the following equation is valid:

[tex]0=v_i+gt_{.5}[/tex]

Also we know that the following equation should be true:

[tex]0=v_i(2t_{.5})+\frac{1}{2}g(2t_{.5})^2[/tex]

Since the kangaroo will end at zero when it hits the ground again.

Solving the system of equations will give you t.

Alternatively, half the time of the jump will be going up and half will be going down.

So

[tex]2.27 m=\frac{1}{2}(t_{.5})^2[/tex]

- Replies
- 9

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 20K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 12K

- Last Post

- Replies
- 12

- Views
- 2K

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 6K

- Replies
- 4

- Views
- 2K

- Replies
- 1

- Views
- 2K

- Replies
- 2

- Views
- 801