4635282

Physics Help

October 16th, 2008 02:13 PM

Someone fires as arrow horizontally over a chasm that is 75m wide. Both sides are the same height. The arrow is released from the shoulder (1.90m above the ground) and has a velocity of 90ms.
How long was the arrow in flight? Where did it land (Distance)

Thanks all.

4635318

Re: Physics Help

October 16th, 2008 02:42 PM

I'm too lazy to get my old school papers out and look, but you will definitely need to clear the angle. 0°?

4635326

Re: Physics Help

October 16th, 2008 02:53 PM

So, you have two equations you need to use to determin the position of the arrow relative to the archer's shoulder as a function of time. One for the horizontal x-axis and one for the vertical y-axis. Assuming our arrow is frictionless this will give you:

x-axis:

y-axis:

Now the first thing we should check is if the arrow will clear the chasm - that is, if the arrow has dropped 1.90 metres, will it have traveled 75 metres horizontally? To do this simply set y as 1.90 and fill in the rest of the formula to find out what t is in this case. Then you take the value you found for t and put it in the formula for the position on the x-axis:

Is x > 75? If it is, then you know where the arrow landed, because it can only have gone down 1.90 metres before it hit the ground and you just calculated what happened in this case. The time traveled is t and the distance traveled is x metres in the x-axis and 1.90 metres in the y-axis.

If that isn't the case, you'll have to go back and fill in x = 75, because in this case, the arrow will not clear the chasm and will hit the chasm wall instead, which is 75 metres away. Once you do that, you do pretty much the same thing you did earlier, except now you calculate t for a certain x value and you use it to find y.

Once you have y, you know that the arrow travelled t seconds and hit the chasm wall at 75m x and ?m y.

4635338

Re: Physics Help

October 16th, 2008 02:57 PM

it's been decades, but you are basicly interested at the time in which gravity has dropped the arrow into ground level. (You are missing tons of variables, so I assume this is for some elementary school physics...)

(delta)t = sqrt(2x(delta)x/a) where x is height of shot and a is gravity (9.8m/s^2), yielding ~0.62sec that translates into 56meters of flight at 90m/s (in frictionless environment).

Assuming bottomless pit, the arrow will hit the opposite cliff therefore giving us flight time of 75m/90m/s ~= 0.8333... sec flight time horizontally.

This can then be used to determine the vertical traverse of the arrow using the first formula (now used as (delta)x = a*(delta)t^2/2). ~= 3.4m. Substracting the starting position of 1.9m (shoulder level? Conan the Destroyer?), it would give us 1.5m below cliff ground level.

This is completely off the top of my head and it's 01:00 am, so this may be incorrect :P

EDIT: Oh well, looks like bunch of people were faster than me ^^

4635344

Re: Physics Help

October 16th, 2008 02:59 PM

Damn i wanted to solve this

4635350

Re: Physics Help

October 16th, 2008 03:04 PM

Someone else got this:
I got the samething.

Hmm, Lightning. I think understand you. Thanks .

4635659

Re: Physics Help

October 16th, 2008 05:59 PM

Do I subtract 1.90 from the second part?

4636794

Re: Physics Help

October 17th, 2008 01:50 PM

lol Coders.
"no prob! Just divide x by y and subtract z while keeping the vector in mind."

I would not even know where to start. I hate math. This is 9th or 10th grade. I remember a similar excercise in a test where a tree should not fall on a house when being cut. My mark in those years was 'D'. (That was really good before I had an 'F' in 11th and 12th grade.)

You're geniuses. Definetely.

4637402

Re: Physics Help

October 17th, 2008 05:11 PM

try to draw out the situation, resolve your initial velocity into x,y components....

drawing situation clearly would help you immensely!

4637520

Re: Physics Help

October 17th, 2008 06:51 PM

thanks all

the answer was t=.833 and falls 1.5m down into the chasm