Vertical Motion Question (physics)

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kmce

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Hi, im not sure if I should be actually asking this in a physics forum, so apologies if this is not suitable for here, but the part I am stuck on is pretty basic for people who actually understand math I think, unlike myself xD.

I am trying to understand a Vertical Motion question, which is

A girl flips a coin into a 50 m deep wishing well. If she flips the coin upwards with an initial velocity of 5 m/s:
At the top of the coin’s flight, the velocity will equal zero. With this information, we have enough to use equation 3 from above to find the position at the top.

v2 = v02 – 2a(y – y0)
0 = (5 m/s)2 + 2(-9.8 m/s2)(y – 0)
0 = 25 m2/s2 – (19.6 m/s2)y
(19.6 m/s2)y = 25 m2/s2
y = 1.28 m

Full question described here https://sciencenotes.org/vertical-motion-example-problem-coin-toss-equations-motion/

Firstly, how does the answer come out to be 1.28m? Wouldnt it be 25 - 19.6?

and secondly, I assume the 0 on the second line is the V02 from the first line?
 
Let's go at this from a consideration of energy (conservation):

[MATH]\Delta E=0[/MATH]
[MATH]\Delta K+\Delta P=0[/MATH]
[MATH]\frac{1}{2}m\left(v_f^2-v_i^2\right)+mg\left(h_f-h_i\right)=0[/MATH]
[MATH]\left(0^2-5^2\right)\left(\frac{\text{m}}{\text{s}}\right)^2+2\left(9.8\frac{\text{m}}{\text{s}^2}\right)\left(h-0\right)\text{ m}=0[/MATH]
[MATH]h=\frac{25}{19.6}\,\text{m}\approx1.28\text{ m}[/MATH]
We could also approach this dynamically:

[MATH]h(t)=-\frac{1}{2}gt^2+v_0t+h_0[/MATH]
[MATH]v(t)=-gt+v_0[/MATH]
Equating the velocity to zero, we find:

[MATH]t=\frac{v_0}{g}[/MATH]
Hence:

[MATH]h\left(\frac{v_0}{g}\right)=-\frac{1}{2}g\left(\frac{v_0}{g}\right)^2+v_0\left(\frac{v_0}{g}\right)+h_0=\frac{v_0^2+2gh_0}{2g}[/MATH]
Plugging in the given data

[MATH]h_{\max}=\frac{25}{19.6}\,\text{m}\approx1.28\text{ m}[/MATH]
Does that help you understand why we divide by twice the acceleration rather than subtract?
 
kmce said:
Hi, im not sure if I should be actually asking this in a physics forum, so apologies if this is not suitable for here, but the part I am stuck on is pretty basic for people who actually understand math I think, unlike myself xD.

I am trying to understand a Vertical Motion question, which is

A girl flips a coin into a 50 m deep wishing well. If she flips the coin upwards with an initial velocity of 5 m/s:
At the top of the coin’s flight, the velocity will equal zero. With this information, we have enough to use equation 3 from above to find the position at the top.

v2 = v02 – 2a(y – y0)
0 = (5 m/s)2 + 2(-9.8 m/s2)(y – 0)
0 = 25 m2/s2 – (19.6 m/s2)y
(19.6 m/s2)y = 25 m2/s2
y = 1.28 m

Full question described here https://sciencenotes.org/vertical-motion-example-problem-coin-toss-equations-motion/

Firstly, how does the answer come out to be 1.28m? Wouldnt it be 25 - 19.6?

and secondly, I assume the 0 on the second line is the V02 from the first line?
Click to expand...
There are two 0s on the 2nd line! Which one are you referring to?
The 1st 0 is v2 which is the velocity at the top. The 2nd 0 is for y0.
 
MarkFL said:
Let's go at this from a consideration of energy (conservation):

[MATH]\Delta E=0[/MATH]
[MATH]\Delta K+\Delta P=0[/MATH]
[MATH]\frac{1}{2}m\left(v_f^2-v_i^2\right)+mg\left(h_f-h_i\right)=0[/MATH]
[MATH]\left(0^2-5^2\right)\left(\frac{\text{m}}{\text{s}}\right)^2+2\left(9.8\frac{\text{m}}{\text{s}^2}\right)\left(h-0\right)\text{ m}=0[/MATH]
[MATH]h=\frac{25}{19.6}\,\text{m}\approx1.28\text{ m}[/MATH]
We could also approach this dynamically:

[MATH]h(t)=-\frac{1}{2}gt^2+v_0t+h_0[/MATH]
[MATH]v(t)=-gt+v_0[/MATH]
Equating the velocity to zero, we find:

[MATH]t=\frac{v_0}{g}[/MATH]
Hence:

[MATH]h\left(\frac{v_0}{g}\right)=-\frac{1}{2}g\left(\frac{v_0}{g}\right)^2+v_0\left(\frac{v_0}{g}\right)+h_0=\frac{v_0^2+2gh_0}{2g}[/MATH]
Plugging in the given data

[MATH]h_{\max}=\frac{25}{19.6}\,\text{m}\approx1.28\text{ m}[/MATH]
Does that help you understand why we divide by twice the acceleration rather than subtract?
Click to expand...

I am not good at math at all, so this will take me some time to go through and fully understand, but just wanted to post just now and say thanks for the reply.
 
Jomo said:
There are two 0s on the 2nd line! Which one are you referring to?
The 1st 0 is v2 which is the velocity at the top. The 2nd 0 is for y0.
Click to expand...
Sorry, what i meant was does the v02 in the first line turn into the 0= (the first 0) in the second line, because (i think) v02 is the starting velocity, so it is equal to 0?
 
kmce said:
Hi, im not sure if I should be actually asking this in a physics forum, so apologies if this is not suitable for here, but the part I am stuck on is pretty basic for people who actually understand math I think, unlike myself xD.

I am trying to understand a Vertical Motion question, which is

A girl flips a coin into a 50 m deep wishing well. If she flips the coin upwards with an initial velocity of 5 m/s:
At the top of the coin’s flight, the velocity will equal zero. With this information, we have enough to use equation 3 from above to find the position at the top.

v2 = v02 – 2a(y – y0)
0 = (5 m/s)2 + 2(-9.8 m/s2)(y – 0)
0 = 25 m2/s2 – (19.6 m/s2)y
(19.6 m/s2)y = 25 m2/s2
y = 1.28 m

Full question described here https://sciencenotes.org/vertical-motion-example-problem-coin-toss-equations-motion/

Firstly, how does the answer come out to be 1.28m? Wouldnt it be 25 - 19.6?

and secondly, I assume the 0 on the second line is the V02 from the first line?
Click to expand...
The units for 25 is (m/s)2 while the units for 19.6 is m/s2. You CAN'T add if the units are different! That is why you should always write down the units!
 
kmce said:
Sorry, what i meant was does the v02 in the first line turn into the 0= (the first 0) in the second line, because (i think) v02 is the starting velocity, so it is equal to 0?
Click to expand...
What about the part of the question that states If she flips the coin upwards with an initial velocity of 5 m/s? The initial velocity is 5m/s or maybe -5m/s.
Also you need to see how nicely things were replaced. The left side initially was v2 and the 2nd line started off at 0. So v2=0
 
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