# Homework Help: One / Two Dimensional Motion

1. Sep 18, 2010

### J.live

3. The attempt at a solution[/b

I tried to use the

X= Vixt + 1/2axt^2 => 0= 1/2axt^2+ Vixt + X

To find the time since there isn't a horizontal distance given.

Then I realized the initial velocity for x and y direction can't be the same , but I don't have the angle.

Do i have to find the angle first?

If so, what do I use ?

Edit: I understand how to solve it now. Initial velocity in y direction is zero so I can solve for the time using the same equation above.

Here is the question

#### Attached Files:

• ###### Screen shot 2010-09-18 at 2.27.11 PM.png
File size:
4.8 KB
Views:
112
Last edited: Sep 18, 2010
2. Sep 18, 2010

### Thaakisfox

It is a horizontal projection so the velocity at the beginning only has an x component. And only the height of the tower is given.
hint: freefall