@@mhmdhojeij2227in vacuum, the weight doesnt affect the dropping speed, and in normal conditions, heavy things like rock can overcome air resistance, so weight doesnt matter
The formula is s=ut+1/2at^2. Since he was holding it still and dropped it with not additional force, u is 0 m/s so the formula becomes s=1/2at^2. t^2 is the time in seconds multiplied by itself. 1/2at^2 is half of the acceleration due to gravity on earth (32 f/s = 16 or 9.8 m/s = about 5). If you multiply those values together you get s which stands for displacement - in this case the height which the rock fell. The mass of the object is irrelevant for this formula although technically it doesn’t account for air resistance, but that is negligible.
@@ridog00 This formula doesn’t take external factors such as air resistance into account. It’s basically an ideal scenario, where you have an initial velocity ‘u’ and you accelerate at the rate ‘a’ for ‘t’ seconds. No additional forces. This isn’t the case in the real world though and different objects may experience different amounts of air resistance, this leading to different outcomes even though the formula doesn’t reflect this.
He forgot to account for the 0.24 second delay (which is the average human reaction time) which would put it at 92 feet deep. Since the drop is so short it’s hard to be accurate; the longer the drop time the more accurate this calculation becomes.
I think the delay between the rock being dropped and the timer starting would be less than .24 seconds because he isn’t reacting to the rock being dropped, he drops it himself
For the people who don’t understand how he calculated that: s=0.5 • a • t^2 s is distance in feet (in this case) a is acceleration in feet per second, the gravitational constant or 32 feet per second t is time in seconds That leaves us with s=16•t^2
@@jumpjump-oz2pr Once the height is great enough air resistance becomes a factor and you have to account for this by using the same equation S = UT + 1/2 AT^2 for that part of the drop. Only acceleration is now zero and starting velocity is the terminal velocity, in this case of the rock.
Since the initial velocity is zero(he dropped it from the rest, a = 9.82 m/s²), the height will be 34.48 meters, which is very close to the height he found: 112.36 feet (34.51 m).
@@mukulsingh3083the speed isn't 9.82 it is the acceleration Speed isn't constant in motion under gravity as there is constant acceleration therefore each second speed increases until it strikes the ground
@@mukulsingh3083There is no 1 "speed" here because its not constant. If you are thinking "velocity = distance / time", thats not accurate here because that only works for constant velocitys when we are assuming its a constant acceloration, not constant velocity, so a linear velocity in respect to time.
For those of who are still wondering how he calculated, here is what he did: On Y axis, gravity is the only acceleration, which is g= 9.8 m/s² (in Feet is g= 32 ft/s² ) And the time he measured was t= 2.65 seconds From Kinematics equations, we can get distance by D= 1/2 g t² _Calculating in Feet_ D = 1/2 × 32 × (2.65)² D = 16 × (2.65)² { This is what he did } D = 112 ft Approx. [34 m] He rounded it up to 110 ft because of time delay he might have made at the drop.
The equation to calculate this is very simple and can be very useful in other ways! X=x0+v0*t+0.5*a*t^2 X is the final position X0 is the initial one V0 is the initial speed T is time A is acceleration In this case we can replace a with 9.8 which is earth's gravity and time with the time it took which was 2.6 seconds, or if you don't have access to the time but you know the height you can switch the equation to look like this: X=0+0+0.5*9.8*t^2 30.48/4.9=t^2 √6.22=t 2.5=t (That's assuming the original height was true) Anyway, thanks for reading this :)
Proof - h = ut + 1/2 gt² h = 1/2gt². ( u = 0 { no initial momentum , hence ut = 0 }) h = 1/2 × 9.8 × (2.65)² h = 4.9 × 2.65 × 2.65 h = 12.985×2.65 h = 34.41025m or 112.895ft ( approx)
@@diy_wizardI can tell u where the 16 came from. I’m studying to be an engineer. The equation is distance= (initial velocity) x (time) + 1/2 x (acceleration) x (time^2). There is 0 initial velocity so we can get rid of that part of the equation. Now he did the time squared so we know that. The 16 comes from when you convert the acceleration due to gravity which is 9.8 meters per second^2 to feet per second^2. There are 3.3 feet to a meter so if you do 3.3 x 9.8 you get 32.34. Now multiply the 1/2 in and you get 16.17. He just rounded it down to 16.
@@creepergaming3445 The SUVAT equation S = UT + (A(T^2))/2 Of course I would still test the jump with a weight similar to my own mass before I did it just to be sure 😆 EDIT: If the rock reaches terminal velocity then you have to use the same equation for that part of the drop, only this time A is zero and U is non-zero so the part of the equation ((A(T^2))/2) disappears instead, leaving S = UT. Multiplying that terminal velocity by the time spent at that velocity yields the distance in metres.
Mark, it's not the most reliable technique because the reaction time you have is not that of an F1 driver, which means that you have a margin of error. To make it more reliable, I would advise putting a sensor on the surface of the water and making sure that as soon as the sensor is activated, the timer stops. After that, you're the engineer.
A small note to people out there: do not try this method for bungee jumping if you don't understand how or why it works. If you mess up with the time measurement, even a little, you will hit the ground. The measly difference between 2.5 seconds and 2.65 seconds is *12 whole feet*. Better just tie a decently heavy rock to a long twine and measure it in some other, more reliable, way To those who wish to know why and how this method works, you need a single easy formula from physics: S(t) = vt + at²/2. It states: the distance (S) an object travels in time (t) equals to the sum of its starting velocity (v) multiplied by time (t) and its acceleration (a) multiplied by time squared (t²), divided by two (/2) In this example, we calculate the distance by specifically dropping the rock, not throwing it, because we need the starting velocity to be 0, otherwise the calculation will be way off. Earlier I said that 0.15 second mistake makes up the difference of 12 feet, now we'd be talking 20+ feet for the same mistake When we measure the time, we need to be quite precise, as that time is later multiplied by itself, which means any mistakes made during the measurement get multiplied as well. Time also must be measured in seconds And finally for the acceleration, if the rock was properly dropped, its acceleration will be equal to the acceleration due to gravity, which equals 9.8 m/s² or 32.2 ft/s², this is where "multiply by 16" bit comes from. We multiply the time squared by acceleration, which is 32.2 in the imperial system, and divide it by two, to get roughly 16
Haha I thought the dumbest comment was the guy who actually thought this worked. Now I read your comment 😂😅 wow the stupidity of people now days. He doesn't care about you or your entertainment. He cares about his wallet.
AP Physics I came in clutch: X = Xi + Vt + 1/2at^2 Simplify a bit since we start at position zero and zero velocity: X = 1/2at^2 We know that acceleration is around 9.8 m/s^2 so we can plug that in: X = 1/2(9.8)t^2 Now, just plug in time for T: X = 1/2(9.8)(2.65)^2 This gets us 34.41 meters or 112.89 feet!
S= gt²/2, g-ускорение свободного падения, S= (9,8*2,65²)/2= 33,3568, но стоит сделать поправку на сопротивления воздуха, в итоге выходит что-то близкое к 30 метрам
the equation was used in this video is S = 1/2at^2 + v0t + h0, because he was stand on the bridge so his v0 is equal 0 and h0 = 0 too, so infer that he just used the equation 1/2at^2
Только эти расчеты не учитывают сопротивление воздуха, время реакции человека, и то, что какое-никакое, но движение кистью было, а значит, была и начальная кинетическая энергия. Также нельзя со 100% уверенностью утверждать, что камень был брошен идеально ровно, и стоит вычесть 1 метр, поскольку бросал ты от перил моста
This would only actually be accurate in a vacuum thought cause the formula he used Distance = (initial velocity × time) + 0.5 × (acceleration × time²) Distance ≈ 0 + 16 × time ² It won't be 16 because of air resistance and because we don't the exact air resistance but the answer won't be far off but factually incorrect
actual metric: equation is derived off of s=ut + 1/2 * a * t^2 u is initial velocity and therefore 0 a is 9.81 as it is acceleration due to gravity t is time taken to fall s is displacement (although here it is just the distance fallen)
Someone mention that average reaction time is 0,24s for this bias in the measure would be 6.52 meters. Please be aware that this bias is rising with the height of the measured distance. As velocity is rising with time.