Motion throughout a parabola

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opus
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Homework Statement


You throw a ball straight upward with initial speed ##v_0##. How long does it take to return to your hand?

Homework Equations


##v=v_0+at##
##x-x_0=v_0t+\frac{1}{2}at^2##

The Attempt at a Solution


My question is a general one that relates to the described solution.
According to the text, the time to reach the top of the flight is ##t=\frac{v_0}{g}##.
It then goes on to say that the downward motion takes the same time as the upward motion, so the total time back to the ground is ##\frac{2v_0}{g}##.
But why would the time to the ground be the same as the time to the top? My confusion lies in the fact that on the way up, we have an initial velocity not equal to zero. On the way down, the initial velocity is zero, so it wouldn't start out going as fast as the first part of the parabola. Then I thought that the first part of the parabola, although it has an initial velocity not equal to zero, does slow down throughout it's entire flight up. In contrast, on the way down, it has an initial velocity of zero but is speeding up until it hits the ground.

So my question is, do these two halves of the parabola always balance out? That is, will the initial velocity of the first half of the flight, be equal to the final velocity of the second half of the flight?
 

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  • #2
scottdave
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Without air resistance, if the path is symmetrical, then yes. Think about this - the path down is like running in reverse of the upward trip,so what will the final velocity be?
 
  • #3
opus
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I want to say that the velocity when it hits the ground will be the same as the velocity when it left the ground, but I can't back that up with any kind of reasoning.
 
  • #4
scottdave
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What about conservation of energy? Do you know about how the energy in the ball changes between potential and kinetic? Think about how much kinetic and potential energy it has at the start, at the top, and at the end.
 
  • #5
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When you toss a ball upward and then catch it, it leaves your hand with a certain velocity and when it returns it has that same velocity but directed downward assuming you catch it at the same height as when it left your hand.

The ball exchanges the kinetic energy of the throw to potential energy at the apex of the flight and then when it strikes your hand it’s exchanged that potential energy back to the same amount of kinetic energy and hence has the same velocity.

It’s a lot like pendulum movement or movement on a swing converting kinetic energy of movement to potential energy and back again until friction saps the energy from the system.
 
  • #6
opus
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Ok thanks guys. We haven't gone over any kind of energy yet. The current section is all on 1-D motion so I'm not sure if I'm supposed to solve the problem using the kinematic equations but what you say makes sense.
 
  • #7
opus
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Here is the solution description.
 

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  • #8
haruspex
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not sure if I'm supposed to solve the problem using the kinematic equations
Kinematics is something else. You mean the 'SUVAT" equations, being the kinetic equations for constant acceleration.
One of these is s=ut+½at2. This equation remains valid as long as the acceleration does not change. There is nothing special about the top of the trajectory. Gravity does not care about that, just keeps on pulling with the same force.
Since we want the time to return to start we can set s=0 and a=-g to get v0t=½gt2. One solution, obviously, is t=0. What is the other?
 
  • #9
opus
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Kinematics is something else. You mean the 'SUVAT" equations, being kinetic equations for constant acceleration.
I'm just calling it what it says in the text. I'd call it what people are familiar with, but this is all I've seen it referred to as. I was just confused as to why we could say the time to the top was the same as the time to the bottom as it was kinda just stated in the text without explanation but was important in solving it symbolically.
 

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  • #10
haruspex
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I'm just calling it what it says in the text. I'd call it what people are familiar with, but this is all I've seen it referred to as. I was just confused as to why we could say the time to the top was the same as the time to the bottom as it was kinda just stated in the text without explanation but was important in solving it symbolically.
Kinematics can be thought of as the geometry of motion. It considers the relationships between velocities and accelerations of parts of a system based on constraints, such as fixed lengths and angles. It is unconcerned with forces and masses. See https://en.m.wikipedia.org/wiki/Kinematics.

The point of my previous post was to show a slightly simpler way of getting to the answer, mostly because I often see students making unnecessary work for themselves by splitting a constant acceleration motion into stages, as here.

With regard to why the descent time is the same, the simplest explanation was given in post #2: the descent is just like the ascent, but with time run backwards.
This can be seen by considering the acceleration equation, ##\frac{d^2y}{dt^2}=-g##. If we flip the sign of t we get ##\frac{d^2y}{(-dt)^2}=-g##, but (-dt)2=dt2, so we are back to the same equation.
 
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  • #11
opus
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Excellent explanation, thank you!
 

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