- #1

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I agree with this logically but I can’t prove it mathematically...

Can you please show me the mathematical proof for this fact without using calculus skills?

Thank you! ^^

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In summary, the time it takes for a vertically thrown ball to reach max height is the same as the time it takes for the same ball to fall from max height to ground level, because the equations for motion with constant acceleration can be used to show that the time and velocity at the highest and lowest points are the same. This proof does not require the use of calculus and can be done with basic kinematic equations.

- #1

- 1

- 0

I agree with this logically but I can’t prove it mathematically...

Can you please show me the mathematical proof for this fact without using calculus skills?

Thank you! ^^

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- #2

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What formulae have you been taught for motion under constant acceleration?oofllama said:

I agree with this logically but I can’t prove it mathematically...

Can you please show me the mathematical proof for this fact without using calculus skills?

Thank you! ^^

- #3

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x = x0 + v0 t + 1/2 a t^2

Plug in the numbers and voila’. However without calculus I’m not sure what to say

I guess one thing to say is that the laws of physics don’t change under time reversal, but that’s a statement, not a proof.

- #4

Homework Helper

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What is the time dependence of the velocity v?Cutter Ketch said:

x = x0 + v0 t + 1/2 a t^2

Plug in the numbers and voila’. However without calculus I’m not sure what to say

I guess one thing to say is that the laws of physics don’t change under time reversal, but that’s a statement, not a proof.

- #5

Gold Member

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We have to use two equations that are offered to us by basic kinematic theory for the case of motion with constant acceleration a.

##v=v_0+at## (1)

##s=v_0t+\frac{1}{2}at^2## (2)

The strict proof of these two equations requires calculus but can be done also with graphical methods, by the graph of acceleration versus time (for the proof of (1)) and by the graph of velocity versus time (for the proof of(2)).

So let's see how we apply these two equations in this problem. The constant acceleration is the gravitational acceleration g (will be changing sign depending how we apply the equations).

When we throw the ball up with initial velocity ##v_0## its velocity ##v## at time ##t## is given by equation (1), where ##a=-g## because now the gravitation acceleration opposes the initial velocity ##v_0##

So it will be ##v=v_0-gt## (3). If we put ##v=0## that means that the ball is at its highest point so the time ##t_u## to reach there can be found if we solve (3) for the time so it will be ##0=v_0-gt_u \Rightarrow t_u=\frac{v_0}{g}##

Equation (2) will become (because in this case acceleration is ##a=-g##) ##s=v_0t-\frac{1}{2}gt^2## . if in this last equation we plug ##t=t_u=\frac{v_0}{g}## and after we do some algebra we ll find that the distance (height) ##s_u## that is traveled for the time duration of ##t_u## is equal to ##s_u=\frac{v_0^2}{2g}##.

So, the time to go up is ##t_u=\frac{v_0}{g}## and the distance (maximum height) that it travels is ##s_u=\frac{v_0^2}{2g}##.

Now we take the case where the ball starts falling from its highest point. We apply again equation (1) but this time acceleration is taken as positive ##a=g##, and also the initial velocity ##v_0=0##. So if we apply equation (1) for the velocity ##v_d## that it will have at the lowest point, and suppose that this will take time ##t_d## we ll have from (1)

##v_d=0+gt_d## (4)

Now we know that the distance that it will travel going down ##s_d## is the same as ##s_u## so it will be ##s_d=s_u=\frac{v_0^2}{2g}## (the ##v_0## here is the ##v_0## from the process of going up).

The time it takes to go down ##t_d## can be found if we apply equation (2) . It will be

##s_d=0+\frac{1}{2}gt_d^2 \Rightarrow t_d=\sqrt\frac{2s_d}{g}##. If we plug ##s_d=\frac{v_0^2}{2g}## and after some algebra we ll find that the time ##t_d=\frac{v_0}{g}##. So it will be ##t_d=t_u## which we wanted to proof.

And also the velocity ##v_d## after it falls at the lowest point will be from (4) ##v_d=gt_d=g\frac{v_0}{g}=v_0##

##v=v_0+at## (1)

##s=v_0t+\frac{1}{2}at^2## (2)

The strict proof of these two equations requires calculus but can be done also with graphical methods, by the graph of acceleration versus time (for the proof of (1)) and by the graph of velocity versus time (for the proof of(2)).

So let's see how we apply these two equations in this problem. The constant acceleration is the gravitational acceleration g (will be changing sign depending how we apply the equations).

When we throw the ball up with initial velocity ##v_0## its velocity ##v## at time ##t## is given by equation (1), where ##a=-g## because now the gravitation acceleration opposes the initial velocity ##v_0##

So it will be ##v=v_0-gt## (3). If we put ##v=0## that means that the ball is at its highest point so the time ##t_u## to reach there can be found if we solve (3) for the time so it will be ##0=v_0-gt_u \Rightarrow t_u=\frac{v_0}{g}##

Equation (2) will become (because in this case acceleration is ##a=-g##) ##s=v_0t-\frac{1}{2}gt^2## . if in this last equation we plug ##t=t_u=\frac{v_0}{g}## and after we do some algebra we ll find that the distance (height) ##s_u## that is traveled for the time duration of ##t_u## is equal to ##s_u=\frac{v_0^2}{2g}##.

So, the time to go up is ##t_u=\frac{v_0}{g}## and the distance (maximum height) that it travels is ##s_u=\frac{v_0^2}{2g}##.

Now we take the case where the ball starts falling from its highest point. We apply again equation (1) but this time acceleration is taken as positive ##a=g##, and also the initial velocity ##v_0=0##. So if we apply equation (1) for the velocity ##v_d## that it will have at the lowest point, and suppose that this will take time ##t_d## we ll have from (1)

##v_d=0+gt_d## (4)

Now we know that the distance that it will travel going down ##s_d## is the same as ##s_u## so it will be ##s_d=s_u=\frac{v_0^2}{2g}## (the ##v_0## here is the ##v_0## from the process of going up).

The time it takes to go down ##t_d## can be found if we apply equation (2) . It will be

##s_d=0+\frac{1}{2}gt_d^2 \Rightarrow t_d=\sqrt\frac{2s_d}{g}##. If we plug ##s_d=\frac{v_0^2}{2g}## and after some algebra we ll find that the time ##t_d=\frac{v_0}{g}##. So it will be ##t_d=t_u## which we wanted to proof.

And also the velocity ##v_d## after it falls at the lowest point will be from (4) ##v_d=gt_d=g\frac{v_0}{g}=v_0##

Last edited:

The fact is that all objects, regardless of their mass, fall at the same rate due to gravity.

Proving this fact is important because it helps us understand the fundamental laws of physics and how the universe works. It also has practical applications, such as in engineering and construction, where the effects of gravity must be accounted for.

This fact can be proven through experiments and observations. One example is dropping objects of different masses from the same height and observing that they hit the ground at the same time.

No, there are no exceptions to this fact. It is a fundamental law of physics that has been proven through numerous experiments and observations.

The implications of this fact are vast and have contributed to our understanding of the universe. It has also led to the development of technologies such as satellites and space travel, which rely on an understanding of gravity and the laws of motion.

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