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You are part of a search-and-rescue mission that has been called out to look for a lost explorer. You've found the missing explorer, but you're separated from him by a 200 m high cliff and a 30 m wide raging river. To save his life, you need to get a 4.4 kg package of emergency supplies across the river. Unfortunately, you can't throw the package hard enough to make it across. Fortunately, you happen to have a 1.8 kg rocket intended for launching flares. Improvising quickly, you attach a sharpened stick to the front of the rocket, so that is will impale itself into the package of supplies, then fire the rocket at ground level toward the supplies. What minimum speed must the rocket have just before impact in order to save the explorer's life?


I know the rocket needs to travel 202.237 m (the hypotenuse of the triangle). So d=202.237. I am lost however, where to go from there.

Any ideas?


Thanks

2006-11-04 05:36:59 · 3 answers · asked by Anonymous in Science & Mathematics Physics

"The rocket travels in a parabolic trajectory (we assume it has a zero burn duration). It is at the peak with 0 vertical velocity when it lands on the cliff edge at s=200 m height. Under 1g deceleration, it takes SQRT(2*s/g) = 6.387 sec to reach this height. The horizontal distance traveled is 30 m in 6.387 sec for a velocity of 4.607 m/sec.
I hope you realize that when you impaled the package with the rocket, all the blood plasma, propane and brandy were lost."

That does not appear to be right, any other thoughts?

Thank you very much.

2006-11-05 08:59:04 · update #1

Sorry, the problem was that the rocket travels from the top of the cliff to the ground below where the explorer is. That is why the value you provided was incorrect.

2006-11-07 00:45:53 · update #2

3 answers

The rocket travels in a parabolic trajectory (we assume it has a zero burn duration). It is at the peak with 0 vertical velocity when it lands on the cliff edge at s=200 m height. Under 1g deceleration, it takes SQRT(2*s/g) = 6.387 sec to reach this height. The horizontal distance traveled is 30 m in 6.387 sec for a velocity of 4.607 m/sec.
I hope you realize that when you impaled the package with the rocket, all the blood plasma, propane and brandy were lost.

EDIT: It would help if you said why it doesn't seem right. But if it isn't right, we must not be trying to get the payload up the cliff, just across the river. Which makes the cliff height another irrelevant parameter, along with the rocket and payload masses, and also means our rocket doen't travel the hypotenuse as you thought. We want a minimum-energy way to fling the stuff across the river (distance S = 30m), which means a 45-degree launch elevation (and still a parabolic trajectory). At launch, vertical velocity Vy = horizontal velocity Vx. To solve for velocity we use two definitions of flight time T: T = 2Vy/g (the time to ascend to peak height and then descend to the ground; of course T also = 2Vx/g), and T = S/Vx (hor. distance/hor. velocity). Then 2Vx^2 = Sg = 30m*9.80665m/s^2, so Vx=Vy=12.128 m/s, and the velocity magnitude = 17.152 m/s. This is the launch and impact velocity.
Of course if the rocket has more energy than this minimum value, one must change the launch elevation to compensate and impact velocity will be greater. If it has less energy than the minimum, the explorer croaks.
Now if this version doesn't seem right, maybe we're on top of the cliff and have to propel it downwards and across the river? I'll let you solve that one.
P.S. Your problem statement is confusing. You say "... then fire the rocket at ground level toward the supplies." Do you mean toward the explorer?

2006-11-04 06:24:07 · answer #1 · answered by kirchwey 7 · 0 0

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2014-09-25 18:30:52 · answer #2 · answered by Anonymous · 0 0

the average speed and angle of the mountain need to be put in the equasion to

2006-11-04 13:39:55 · answer #3 · answered by Andrew P 2 · 0 0

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