- #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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- #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!! ^^

- #2

- 35,878

- 6,598

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

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

- 950

- 418

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

ehild

Homework Helper

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What is the time dependence of the velocity v?

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

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- 1,385

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 lets 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 travelled 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 lets 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 travelled 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##

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