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 July 10th, 2019, 07:11 AM #1 Senior Member   Joined: Aug 2014 From: India Posts: 470 Thanks: 1 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 & $\displaystyle Y_{max}$? July 10th, 2019, 12:08 PM   #2
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 Originally Posted by Ganesh Ujwal 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. July 10th, 2019, 07:40 PM   #3
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 Originally Posted by topsquark 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 $\displaystyle y$, $\displaystyle y_{0}$,$\displaystyle v_{0y}$,$\displaystyle V$ & $\displaystyle H$? July 10th, 2019, 08:30 PM   #4
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 Originally Posted by Ganesh Ujwal 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 July 10th, 2019, 09:29 PM #5 Senior Member   Joined: Aug 2014 From: India Posts: 470 Thanks: 1 Still can't figure out formulae of $\displaystyle Y_{max}$. So what is the final formulae for $\displaystyle Y_{max}$? July 11th, 2019, 09:23 AM   #6
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 Originally Posted by Ganesh Ujwal 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 Tags formula, height, max, proper, trajectory, ymax Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post prashantak Physics 4 March 30th, 2019 03:35 PM agentredlum Number Theory 0 May 4th, 2017 10:35 AM triplekite Calculus 2 October 19th, 2012 11:58 PM ZardoZ Applied Math 17 November 17th, 2011 02:01 PM e81 Algebra 4 May 18th, 2011 08:41 PM

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