Kinematics in 1D. I with setting up the question

[hey123a]

Homework Statement

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 6.00m. The stones are thrown with the same speed of 9.0m/s. Find the location (above the base of the cliff) of the point where the stones cross paths.

h = 6.00m
v = 9.0m/s

The answer to the question is
h = 2.45m
t = 1/3

Homework Equations

y = Vot + 1/2at^2

The Attempt at a Solution

Yup = Vot + 1/2at^2
Yup = 9.0t + 1/2(-9.8)t^2
Yup = 9.0t - 4.9t^2

Ydown = h - (Vot + 1/2at^2)
Ydown = 6 - [(-9.0)t + 1/2(-9.8)t^2)]
Ydown = 6 - [-9.0t - 4.9t^2]
Ydown = 6 + 9.0t + 4.9t^2

Cross paths when:
Yup = Ydown
9.0t - 4.9t^2 = 6 + 9.0t + 4.9t^2
-4.9t^2 = 6 + 4.9t^2
-6 = 9.8t^2
t = 0.24999, this is already wrong since the answer for t = 1/3

But I get the right answer for time when I solve this way:
Yup = Ydown
9.0t - 4.9t^2 = 6 - [-9.0t - 4.9t^2]
9.0t = 6-(-9.0t)
18t = 6-0
18t = 6
t = 6/18
t = 1/3

However, when I go to find the height I get different answers:
Ydown = 6 - [-9.0t - 4.9t^2]
Ydown = 6 - [-9.0(1/3) - 4.9(1/3^2)]
Ydown = 3.54444

Yup = 9.0t - 4.9t^2
Yup = 9.0(1/3) - 4.9(1/3^2)
Yup = 2.456

I need help understanding what I'm doing wrong. I've learned that positive values are always objects that move towards the positive x-axis or upwards. And I've learned that negative values are always objects that move towards the negative x-axis or downwards

 

[voko]

Your expression for $Y_{\text{down}}$ is not correct. The correct expression is $Y_{\text{down}} = h + v_0 t + \frac 1 2 t^2$. Then you substitute the values for $h$, $v_0$ and $a$ into it, minding their signs.

 

[hey123a]

Your expression for $Y_{\text{down}}$ is not correct. The correct expression is $Y_{\text{down}} = h + v_0 t + \frac 1 2 t^2$. Then you substitute the values for $h$, $v_0$ and $a$ into it, minding their signs.

Why is it h+ and not h-?
Shouldn't it be h- because the value of Ydown must be less than 6?

 

[voko]

The general equation for displacement with uniform acceleration is $s = s_0 + v_0 t + \frac 1 2 a t^2$. No minus signs in it.

When you apply it to a problem, certain constants in the equation may have minus signs. For example, one frequently has $a = -g$, so instead of $+ \frac 1 2 a t^2$ one gets $- \frac 1 2 g t^2$. Likewise, if initial velocity is downward, $v_0$ becomes $-v_0$. But it is important to be consistent and not apply minus signs twice, like you did.

Don't try to be smarter than it needs. Start with the general equation. Plug in the constants. Mind their signs. Solve.

 

[hey123a]

The general equation for displacement with uniform acceleration is $s = s_0 + v_0 t + \frac 1 2 a t^2$. No minus signs in it.

When you apply it to a problem, certain constants in the equation may have minus signs. For example, one frequently has $a = -g$, so instead of $+ \frac 1 2 a t^2$ one gets $- \frac 1 2 g t^2$. Likewise, if initial velocity is downward, $v_0$ becomes $-v_0$. But it is important to be consistent and not apply minus signs twice, like you did.

Don't try to be smarter than it needs. Start with the general equation. Plug in the constants. Mind their signs. Solve.

Ahh thank you!