My Math Forum What is the proper formula for Max height & Ymax for this trajectory?

 Physics Physics Forum

 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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,267
Thanks: 934

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 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
Senior Member

Joined: Aug 2014
From: India

Posts: 470
Thanks: 1

Quote:
 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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,267
Thanks: 934

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,267
Thanks: 934

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top