Sf=1/2a🔺️t^2+Vo🔺️t+So No string, 1/2(-32)(7.5^2)+(330)(7.5)+0 =1575 ft With string, 1/2(-32)(3.5^2)+(330)(3.5)+0 =959 ft So=starting position, 0 a=negative acceleration of gravity, -32 ft/s (using ft/s because that's what the bow's speed is measured in) Vo=staring velocity, 330 ft/s (we'll just take Edwins word here) 🔺️t=change in time, 15s and 7s (we'll take half of each time because we only want to determine the position of the arrow halfway through its journey when it will be at its highest point) Sf=final position, the highest point in this case Assumptions: 1. Bow is indeed 330 ft/s when shooting that specific arrow. 2. Wind resistance is insignificant due to the aerodynamics of the arrow. 3. Edwin shot both arrows at the exact same angle from the exact same height.
Yeah if its on the tip it would turn the arrow around since the string is pulling the tip down. Atleast I think so. But attaching it to the end of the arrow would be better anyway
In the news, two men found dead in the desert. It seems they were killed by arrows coming from above. One of the men, had blisters on his fingers, likely caused by picking up very hot items.
The string exerts a lot of aerodynamic force on the arrow, so having it on the front is like adding fins on the front (bad for stability). You can actually make darts or arrows without fins using only a string at the rear for stability.
So if it spent 15 seconds in the air, half the time rising half the time falling gravity is 9,8 meters per second per second so after 7.5 second of falling it will be at 73.5 meters a second peak velocity which should equal its starting velocity so multiply that by 7.5 seconds and divide by 2 and you have a height of 275.6 meters.
@@ruebenmikoch1828 The main thing I was not taking into account was air resistance. In a vacuum, the rise and fall are mirror images of each over so I was taking into account the deceleration in the rise. He is shooting pretty close to vertical so lateral movement is not a big factor.
la cuerda sujeta en la punta hará que la flecha se vaya de lado y sea menos aerodinámica, si la ponen en la parte trasera de la flecha, esta se mantiene en la misma dirección....
Eliseo Rubio. But he was almost right in ft he said 200 it was 180 but I thought that was funny too how he was like 100 yards 98 at least then Edwin was like in FT ohh 200 lol
I think it's enough if it's fun, when you're the first to have fun doing these experiments. There is no need to do the calculations just to hear you call cabron against each other. Ciao from Venezia.
I was just scrolling through RU-vid when I was bored and I just watched one video from this channel and it was very funny then I started watching more and more videos and I rarely like videos on RU-vid but this one is was special since then I subscribed to this channel and it been really fun guys Edwin and and gasper you guys are really natural and friendly keep the good work up guys 👍🏼👍🏼❤️
15 seconds travel time, assuming the velocity of which the arrow hits the ground is the same as when it was launched, it would reach the highest point at 7.5 seconds. 9.81*7.5=73.5m/s at launch, with constant acceleration it would have traveled (0+73.5)/2*7.5=276 meters However this doesn't take into account air resistance, so the actual speed would be faster If calculate by the bow's launching speed of 330ft/s, or 101m/s as following: When arrow is at it's highest point, distance from ground is (0+101)/2*7.5=416 meters So with 0 air resistance arrow would reach 416 meters So there's a 140 meters of gap of which the arrow actually could have reached, from 276m to 416m, and it is more likely to be at the middle than the edge of the range If we take the middle, 346 meters from ground, as the base to calculate the average speed air resistance caused to reduce (of cause air resistance slow more at higher speed, but we are taking the total average) (average velocity)*(time spend upward)=distance (101+0)/2*t=346 t=6.23 acceleration*time=change in velocity (9.81+x)*6.23=101-0 x=6.4m/s² During downward motion: t=15-6.23=8.77s acceleration*time*time=distance traveled (9.81-y)*8.77²=346 9.81-y=4.5 y=-5.31m/s² So air resistance caused the arrow to slow 6.4m/s² on average flying upward and 5.31m/s² flying downward With string, assuming the max height reading is correct: Max height=180ft=54.86 meters t=7 average speed*time=distance (101+0)/2*t=54.86 t=0.99s acceleration*time*time=distance traveled (9.81+i)*0.99²=54.86 9.81+i=56 i=46.2m/s the string and air resistance caused the arrow to slow on average at 46.2m/s² Downward: t=7-0.99=6.01s average speed*time=distance (v+0)/2*6.01=54.86 v=18.3m/s the arrow hit the ground at 18.3m/s acceleration*time*time=distance traveled (9.81-j)*6.01²=54.86 9.81-j=1.52 j=8.3m/s² So air resistance and the string caused the arrow to slow on average 46.2m/s² upward and 8.3m/s² downward Air resistance in first cast is greater in second case, if we plug the number from first into second: The string caused the arrow to slow
Haha 😂😂 No doubt about it, man. My wife is always telling me I'm going to die before she does because I'm always doing stuff that seems normal to me, but she says those things I see as "normal" are highly dangerous. 😂😂😂 I just don't see how the things we do are dangerous 🤷♂️🤷♂️ We just have a different nature than women do, and that's all I can say.
It went up about 1102.5 meters. We can't know exactly how far it went up without knowing a dozen other variables, but we can use the basic free fall formula to get a reasonable estimate of the distance from the time it started falling. Height = 0.5*g*t^2 So, 0.5*9.8*15^2, which comes out to be 1102.5.
Wait Cabrona. Edwin and Gasperito shooting arrows straight up then standing there looking up. Priceless you guys are great.
4 года назад
The line adds a lot of drag. And we need more data. Arrow's velocity at launch, arrows weight, wind velocity (ideal would be little or no wind at all).
Before you start the video you guys should have put rolls of sting on the ground like a lot of loose rolls but all the same you guys are natural I like it 😂😂
You should have stuck a arrow in the ground then put that spool on the arrow shaft. Like put the arrow threw that hole in the spool. Leave the arrow sticking straight up in the air. The string will come off much smoother.
not only gasparito.... many many peoples from south america and even spain say that all the time :D. my dad was spanish, and he say "cabron" sooo many time in his life :D
You 2 are killing us....release that arrow already.....AND RUN FOR COVER! This one was tough to watch for sooooo many reasons, i loved it, but it was pure torture
A new Olympic sport in the making. Not only distance but accuracy. Shooting an apple off someone's head and retrieving the apple as well. Points deducted if you miss and also based on where you shoot them in the head. Great video. 😎 👍👍👍👍👍
Assuming the arrow took 15 seconds to fall without any string attached, it would have traveled about 3600 feet. If the arrow took 7 seconds with the strings attached, the arrow travelled about 784 feet. You can use the formula, 16(x^2) to approximate the height distance of an object from its peak in the air until it touches the ground. The variable x represent the time when taken in seconds. (This doesn’t account for air resistance or the added weight from the string)
It took 15 seconds for the first arrow to make a round trip so it took almost 7.5 seconds to make it to top and assuming that arroe shoots at 330 ft/s by second equation of motion height =vi*t-1/2*g*t*t that comes out to be 1577 ft high!!
@@manuelcastrou79 Manuel Castro But the time taken will always be the same because throughout this whole trip the only vertical firce acting is the gravity which is downwards (the second derivative of height). The speed does surely vary all around but the equation only requires the initial speed and the angle to be exactly 90 degrees..
But the velocity is slowing the moment it leaves the bow until it stops at the highest point & the wind is gusty verying the drag on the line. Besides, the whistling head is both wider than normal target/feild heads & the holes at as mini-air-brakes. I think all their method really tells us is how high that arrow goes given the number of glaring faws in their set-up.
New drinking game....take a drink every time either one of them says “CABRON!!” 🤢🤮🤮🤮 P.S. Somebody needs to start keeping a “CABRON” count on each video 👍😉
ok guys The body of the M-67 hand grenade is a 2.5-inch diameter steel sphere designed to burst into numerous fragments when detonated. It produces casualties within an effective range of 49.5 yards (15 meters) by the high velocity projection of fragments. The grenade body contains 6.5 ounces of high explosive. Each grenade is fitted with a fuse that activates the explosive charge. Some fragments can go far as 230 meters!
What you could do is put a light washer around the string. Shoot it up, and the washer will get shifted to the bottom of the string when tha arrow reaches its peak. And you'd get a rough estimate of how high it went by measuring the distance from the washer to the arrow.
TOW missiles and wire guided torpedoes use a line coiled very much like a roll of kite string unspooled from the center. This provides very little resistance to unspooling and very little drag on your arrow. If you achieved 15 seconds with just the arrow, I would expect 13 seconds with kite string.
4:58 hahahahaha! Used to do this with my friends in middle school. Called it arrow roulette... we all 3 shot an arrow straight up and the last one to move won..
"hahahahaha! Used to do this with my friends in middle school. Called it arrow roulette... we all 3 shot an arrow straight up and the last one to move won.". We played the same game many times. The only difference was that Barry - who hands down was the most committed and thus became the greatest player among us all; after he won, we all decided to retire that game, and move on to other games. Why? Well, without Barry's presence the game just wouldn't be the same. R.I.P. Barry. No one would ever beat his record...no one wanted to try to either.
