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You see, I have to build a device to project as many marbles as possible into a basin 12 in in diameter and 4 in high, that is placed 10 ft away from me. There is a limit of 60 seconds for me. I can only use one marble at a time, the device cant be hand-held. Now, I bought a Hot Wheesls V Drop Super Velocity Set: http://www.walmart.com/catalog/detail.gsp?image=http://i.walmart.com/i/p/00/02/70/84/30/0002708430781_AV_60X60.gif,http://i.walmart.com/i/p/00/02/70/84/30/0002708430781_500X500.jpg,http://i.walmart.com/i/p/00/02/70/84/30/0002708430781_AV_60X60.gif,http://i.walmart.com/i/p/00/02/70/84/30/0002708430781_AV1_60X60.gif&product_id=5691446&iIndex=2&isVariant=false&corpCard=false&type=0, and I used its track as a track for my marbles. The only problem is that I cannot make my marbles go 10 ft! AND, I tried every combination possible with height and weight of the track. Help or tips are welcome.

2007-10-05 14:26:55 · 2 answers · asked by inna2507 2 in Science & Mathematics Physics

2 answers

You will get the most distance from the trajectory if the ball leaves the track at 45º from the horizontal. You need to calculate how high the starting point of the ball's drop must be. Let the ball's exit velocity be V; the horizontal and vertical components of that will be V/√2. If the flight time is T, the horizontal distance X traveled will be T*V/√2. To find T, note that the rise and fall times will both be given by g*t = V/√2, so T = 2*V / g√2. This gives X = (V^2)/g; then you need the ball's exit velocity to be V = √[g*X] (X is 10ft).

How much must the ball drop (H) to get that? The potential energy of the ball at start is m*g*H; the kinetic energy when it leaves is 0.5*m*V^2. These must be the same, so

0.5*m*V^2 = m*g*H

H = 0.5*V^2/g; from above, V= √[g*X]

H = 0.5*X.

Thus your starting altitude must be half the horizontal distance, or 5 ft above the floor. Since this does not take into account friction, you may need a little more than that, so you will have to run tests.

In summary, you will need over 5 ft drop and a 45º exit angle.

2007-10-05 14:53:54 · answer #1 · answered by gp4rts 7 · 1 0

i hate this kind of stuff

2007-10-05 14:34:11 · answer #2 · answered by Anonymous · 0 1

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