Determine Initial Velocity, Max Ht for Diver Proj. Motion HW

[TheShehanigan]

Homework Statement

A high diver leaves the end pf a 5.0-m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Determine his inicial velocity, his maximum height and the velocity with which he enters water. (Projectile Motion).

*See details at the end*

Homework Equations

y = yo + volt + 1/2gt^2

The Attempt at a Solution

I have worked on this exercise, and can fully answer it using the water as the origin with the y-axis pointing upward. However, I wanted to try it with the diver leaving the board as the origin with the y-axis pointing downward (thus making g positive) and it's not giving me the correct result. Once I clear y = yo + volt + 1/2gt^2 for vo, I obtain v_o = (y - yo + 1/2gt^2)/t. Since I am using the y-axis downward, I am making y = 5.00m and g = 9.8 m/s^2. I left the +1/2 gt^2 0 upon clearing the formula for vo since the acceleration is always downward (thus positive always).

However, it's not giving me the right result. Someone see anything wrong with my logic?

*I don't need help with the problem per-se, but with why the y component of velocity isn't giving me the right number. *

The correct answer should be 2.5 i for the y component of vo.

 

[LowlyPion]

When you choose down as positive y then the sense of velocity and acceleration are both changed. Picking a proper origin is also complicating.

It should all work out the same, but you will need to take care in what is y=0 and with +/- for V_yo as well as g.

 

[TheShehanigan]

So a = - g for upward becomes a = g for downward, (with g = 9.80 m/s^2) and what would be vo = + for upward becomes vo = - for downward, or what? Also, y = 0 is the exact last point of the diving board in the way I'm trying to make the exercise.

I uploaded a diagram for reference.

 

[LowlyPion]

If V_o is against the direction you chose as positive, then of course it is negative.