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July 10th, 2019, 07:11 AM   #1
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What is the proper formula for Max height & Ymax for this trajectory?

Trajectory:



Time it takes from A to travel to B; $\displaystyle T = \large\frac{Vsinθ}{g}$

Time it takes for for B to travel to C; $\displaystyle T = \large\sqrt\frac{2y_{max}}{g}$

Then what is the formula for Max height & $\displaystyle Y_{max}$?
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July 10th, 2019, 12:08 PM   #2
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Trajectory:



Time it takes from A to travel to B; $\displaystyle T = \large\frac{Vsinθ}{g}$

Time it takes for for B to travel to C; $\displaystyle T = \large\sqrt\frac{2y_{max}}{g}$

Then what is the formula for Max height & $\displaystyle Y_{max}$?
The equation of motion is $\displaystyle y = y_0 + v_{0y} t - (1/2)gt^2$. Here you have $\displaystyle y_0 = H$, $\displaystyle v_{0y} = V ~ sin( \theta )$ and t = T.

You really shouldn't use T to denote two different times. And, personally, I'd change the label V to $\displaystyle V_0$ or something.

-Dan

Last edited by topsquark; July 10th, 2019 at 12:13 PM.
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July 10th, 2019, 07:40 PM   #3
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The equation of motion is $\displaystyle y = y_0 + v_{0y} t - (1/2)gt^2$. Here you have $\displaystyle y_0 = H$, $\displaystyle v_{0y} = V ~ sin( \theta )$ and t = T.
-Dan
What is $\displaystyle y$, $\displaystyle y_{0}$,$\displaystyle v_{0y}$,$\displaystyle V$ & $\displaystyle H$?
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July 10th, 2019, 08:30 PM   #4
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What is $\displaystyle y$, $\displaystyle y_{0}$,$\displaystyle v_{0y}$,$\displaystyle V$ & $\displaystyle H$?
I have put a vertical origin at the base of the cliff. $\displaystyle y_0$ is the initial position, so $\displaystyle y_0 = H$. y is the vertical position, y(t). $\displaystyle v_{0y}$ is the vertical component of the initial speed, $\displaystyle V ~ sin( \theta )$. For some reason you are using the initial speed as "V"... I don't like the notation. And finally, H is the height of the cliff.

-Dan
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July 10th, 2019, 09:29 PM   #5
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Still can't figure out formulae of $\displaystyle Y_{max}$. So what is the final formulae for $\displaystyle Y_{max}$?
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July 11th, 2019, 09:23 AM   #6
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Still can't figure out formulae of $\displaystyle Y_{max}$. So what is the final formulae for $\displaystyle Y_{max}$?
Please tell me you are reasonably familiar with the equations
$\displaystyle s - s_0 = vt + \dfrac{1}{2}at^2$
$\displaystyle v = v_0 + at$
$\displaystyle s - s_0 = \dfrac{1}{2} ( v_0 + v) t$
$\displaystyle v^2 = v_0 ^2 + 2a(s - s_0)$
Where s is the displacement in a given direction. (I have changed to s instead of y because most texts start out that way.)

These are the equations of motion of an object with a constant acceleration. In this case a = -g.

In this case let's look at the motion in the y direction. I'm going to set an origin at the bottom of the cliff directly below where the object was launched, and +y is upward.

So we know at what height the object was thrown (or whatever): $\displaystyle s_0 = H$ and it was launched at a speed V at an angle $\displaystyle \theta $ above the horizontal: so $\displaystyle v_0 = V ~ sin( \theta )$. You are looking for the max height, $\displaystyle s_{max}$, which is where the vertical component of the velocity $\displaystyle v = 0$.

So what equation(s) do we have where we know the values of a, $\displaystyle s_0$, $\displaystyle v_0$, and v and we are looking to find s? There's only one of them in the list.

-Dan
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