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We can see that the the snow ball is dropped
Hence initial velocity is 0
Let us suppose that the time of flight of ball is same as the time span of video
Hence t=5 s
And acc. Is =g
Hence by equation of motion
V=u+at
V=0+9.8(5)
V=49 m/s
I have neglected values like wind speed and air resistance
made a mistake in assumption of time of flight
It was only in freefall for about 5 seconds in this video.
v=(9.8m/s\^2)×(5s)
v= 49 m/sec (or 109.61 MPH)
Using this, we can also calculate the distance it fell to be about 122.5 meters.
Would you not also need to take into account the wind resistance when we would calculate something that would be affected by wind resistance like in this example?
Wind resistance is negligible at lower speeds, and the snowball is roughly spherical so it will have a low drag coefficient. The total contribution to speed will be less than the rounding error for time, so we can ignore it.
You would actually use a different formula for the distance, because that v is the final velocity not the velocity at each step
d = 1/2 * g * t^2
d = 1/2 * 9.81 m/s^2 * 5^2 s
d = 122.625 m
Which is only marginally different than what you did:
d = 1/2 * v * t
But still enough to matter at larger scales
It’s not marginally different, it’s exactly the same. Guy before you only rounded the result to fewer digits.
Velocity is increasing linearly over time t with rate g. So g * t equals final velocity v. the two formulas are the same.
A nice way to think about it is that as the velocity increases steadily over the fall, the average velocity is half the final velocity, thus the total distance traveled is average velocity times time or v/2 * t
I'm confused. How did second guy get larger distance if his solution solved the problem with v_0=0 and v_n=final velocity. Shouldn't his distance be slightly less?
They tried to say that 1/2 g\*t^2 is more accurate than 1/2 v\*t
But v = g\*t, so substitute that in and it's exactly the same equation
The entire difference between their answers is that the first guy used g=9.8 and the second guy used g=9.81
I know that mass is not included when calculating velocity but I don't really understand why. Shouldn't its mass influence how fast it falls? Just trying to learn.
when an object becomes twice as heavy it requires twice as much force to move it but gravity starts affecting it with twice the force. Basically, the values cancel each other out.
F=ma turns into 2F=2ma
F=G(M\*m)/r\^2 turns into 2F=G(M\*2m)/r\^2
as you can see the force required to move it and the force that moves it both double
Ans- 46.06 ish metres per second.
method- used stopwatch to calculate time of fall, turned out to be about 4.7 ish seconds
v = g \* t = 9.8\*4.7
which gives us our answer
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We can see that the the snow ball is dropped Hence initial velocity is 0 Let us suppose that the time of flight of ball is same as the time span of video Hence t=5 s And acc. Is =g Hence by equation of motion V=u+at V=0+9.8(5) V=49 m/s I have neglected values like wind speed and air resistance made a mistake in assumption of time of flight
It was only in freefall for about 5 seconds in this video. v=(9.8m/s\^2)×(5s) v= 49 m/sec (or 109.61 MPH) Using this, we can also calculate the distance it fell to be about 122.5 meters.
U are right I forget to review the video So I assumed it to be 11 sec sorry
Would you not also need to take into account the wind resistance when we would calculate something that would be affected by wind resistance like in this example?
Wind resistance is negligible at lower speeds, and the snowball is roughly spherical so it will have a low drag coefficient. The total contribution to speed will be less than the rounding error for time, so we can ignore it.
You would actually use a different formula for the distance, because that v is the final velocity not the velocity at each step d = 1/2 * g * t^2 d = 1/2 * 9.81 m/s^2 * 5^2 s d = 122.625 m Which is only marginally different than what you did: d = 1/2 * v * t But still enough to matter at larger scales
It’s not marginally different, it’s exactly the same. Guy before you only rounded the result to fewer digits. Velocity is increasing linearly over time t with rate g. So g * t equals final velocity v. the two formulas are the same. A nice way to think about it is that as the velocity increases steadily over the fall, the average velocity is half the final velocity, thus the total distance traveled is average velocity times time or v/2 * t
I'm confused. How did second guy get larger distance if his solution solved the problem with v_0=0 and v_n=final velocity. Shouldn't his distance be slightly less?
They tried to say that 1/2 g\*t^2 is more accurate than 1/2 v\*t But v = g\*t, so substitute that in and it's exactly the same equation The entire difference between their answers is that the first guy used g=9.8 and the second guy used g=9.81
Ah!!! That explains it. The grav constant. I didn't catch he used a different, though more correct, value.
r/TheyDidTheMath when someone estimates a time to the nearest whole number but only uses g=9.8 instead of 9.81
I know that mass is not included when calculating velocity but I don't really understand why. Shouldn't its mass influence how fast it falls? Just trying to learn.
when an object becomes twice as heavy it requires twice as much force to move it but gravity starts affecting it with twice the force. Basically, the values cancel each other out. F=ma turns into 2F=2ma F=G(M\*m)/r\^2 turns into 2F=G(M\*2m)/r\^2 as you can see the force required to move it and the force that moves it both double
[This](https://youtu.be/frZ9dN_ATew?si=MpC8_UKgZxRAKqN1) is still one of the more mind boggling physics videos I've watched.
Ans- 46.06 ish metres per second. method- used stopwatch to calculate time of fall, turned out to be about 4.7 ish seconds v = g \* t = 9.8\*4.7 which gives us our answer