TBR Physics Passage 1 Question 7

[Lacour]

---

Members do not see this ad.

So, the context is that Galileo is throwing a stone of 1 kg horizontally from the top of a 20-foot rocket. He preforms two trials on Earth and two on a planet with half the gravitational force of Earth, but the same atmospheric composition. The results were:

Trial/g [m/s^2]/Vo [m/s]
Stone A/9.8/10.0
Stone B/9.8/5.0
Stone C/4.9/10.0
Stone D/4.9/5.0

#7 asks "how should the ranges of Stone A and Stone D compare" with the answer being A. I agree with the answer, but I'm not sure how they answer key is doing it's math in the solutions guide. It says "Let's look at the formula for range: R = Vox(sqrt(2h/g)) as it applies to trials A and D. Ra = Vox(sqrt(2h/gA)), where VoA = 10 m/s and gA = 9.8 m/s^2 and Rd = Vox(sqrt(2h/gD)), where VoD = 5 m/s and gD = 4.9. Note that VoA = 2VoD and gA = 2gD, so we can write: Ra = 2VoD(sqrt(2h/2gD)) = sqrt(2)*VoD(sqrt(2h/gD)) = (sqrt(2))Rd, which is greater than Rd."

The bolded part is where I lost them. When I plug in the actual numbers and solve, I agree with their answer. I'm just not sure how they're going from the first bolded equation to the second one mathematically.

Help = Awesome, thank you!

 

[whiteshadodw]

So, the context is that Galileo is throwing a stone of 1 kg horizontally from the top of a 20-foot rocket. He preforms two trials on Earth and two on a planet with half the gravitational force of Earth, but the same atmospheric composition. The results were:

Trial/g [m/s^2]/Vo [m/s]
Stone A/9.8/10.0
Stone B/9.8/5.0
Stone C/4.9/10.0
Stone D/4.9/5.0

#7 asks "how should the ranges of Stone A and Stone D compare" with the answer being A. I agree with the answer, but I'm not sure how they answer key is doing it's math in the solutions guide. It says "Let's look at the formula for range: R = Vox(sqrt(2h/g)) as it applies to trials A and D. Ra = Vox(sqrt(2h/gA)), where VoA = 10 m/s and gA = 9.8 m/s^2 and Rd = Vox(sqrt(2h/gD)), where VoD = 5 m/s and gD = 4.9. Note that VoA = 2VoD and gA = 2gD, so we can write: Ra = 2VoD(sqrt(2h/2gD)) = sqrt(2)*VoD(sqrt(2h/gD)) = (sqrt(2))Rd, which is greater than Rd."

The bolded part is where I lost them. When I plug in the actual numbers and solve, I agree with their answer. I'm just not sure how they're going from the first bolded equation to the second one mathematically.

Help = Awesome, thank you!

its a long and tedious derivation. some of their answers require too much tedious derivation and if you were to do this on the test you could take up a good 10 minutes trying to figure it out. usually when you get to a question like this in TBR that is too hard to calculate you should think your way through the problem. the answer is right in front of you and you can start by getting rid of the obviously wrong answers. there will always be at least one, maybe two. then it becomes an issue of finding the "BEST" answer.

the longer a stone stays in air, the longer it can travel horizontally and the larger its range will be. right? ok, so using displacement = (1/2)g*t^2 you can find how long the A and D will be in the air. i know A will be in air longer because of less gravity. but i don't know how much longer. if stone D is in air more than two times longer than A then i know D will travel further despite having a horizontal velocity of only half of stone A. incidentally stone A is in air for about 2 seconds, and D is in air for a little under three seconds, but over 2 seconds. 2 seconds * 10 (m/s) is the range for stone A, ~3 m/s * 5 m/s is the range for stone D. stone A range (20 m) >>> stone D (~15 m this is an upper estimate so we can assume its definitely less than the range of A)

 

[Lacour]

Thanks a ton. Your explanation makes a lot more sense then BRs attempt at manipulating equations.