TBR physics help

[watchntv]

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Im doing Chapter 1, passage 1,
Galileo
I'm lost about the answer, the passage is about projectile motion and the question asks ,to compare flight times
it is comparing 2 horizontal throws, in 2 different gravities(10 and 5)
both objects are thrown at the same rate, the only different is A has a gravity of 10 and C has a gravity of 5
so I thought the answer was A's flight time was half that of C's fight time
since A has double the gravity
but the answer is A has .7*C's flight time...
and the explanation is beyond my understanding, can someone dumb it down and explain this to me?

also, I only figured this out with my calculator, apparently I need to know 1/(square root of 2)?
or what the square root of 2 is?

I do occasionally ask my MD to give me the answer to some basic physics questions, like:
-how much Vit d do I need?
-should I take lipitor?
if I throw a rock off a cliff at 10m/s, how far does the rock travel in the X direction if the cliff is 200meters high
-how do I increase my HDL?

thats a typical visit, right?

 

[watchntv]

nevermind, weight doesnt matter, nor wil gravity

i need coffee

 

[SN2ed]

Thread moved. These types of questions belong in the Study Q&A forum.

 

[ljc]

Im doing Chapter 1, passage 1,
Galileo
I'm lost about the answer, the passage is about projectile motion and the question asks ,to compare flight times
it is comparing 2 horizontal throws, in 2 different gravities(10 and 5)
both objects are thrown at the same rate, the only different is A has a gravity of 10 and C has a gravity of 5
so I thought the answer was A's flight time was half that of C's fight time
since A has double the gravity
but the answer is A has .7*C's flight time...
and the explanation is beyond my understanding, can someone dumb it down and explain this to me?

Horizontal velocity is irrelevant to time of flight. The only thing that matters is initial vertical velocity and gravity. In this case, since you make no mention of the vertical velocity, we should assume they both start from vertical rest. We can also assume they start from the same height. This means object A is accelerating downwards at 10m/s^2, and object C is accelerating downwards at 5m/s^2.

Now we have our picture, and we need to figure out the numbers. I will start by saying that finding time of flight from distance is probably the hardest of the kinematics problems to work through, as it eliminates several of your shortcuts. The quickest way I've found is to just do a little guess and check, then round.

For instance, pretend both objects start at vertical position of 5meters. This means object A has a time of flight of 1 second (0m/s initial, 10m/s final, average velocity 5m/s, time of flight 1s, 5m/s*1s=5m). Now do some guess and check to compare object C: at 2s flight time (your guess), initial velocity is 0m/s, final velocity is 10m/s, average velocity is 5m/s, time of flight is 2s, so 5m/s*2s=10m. So object C falls 10m in 2s -- this is too much, so it must be less than that. Let's try 1.4 (the square root of 2, a very common number in physics problems): initial velocity is 0, final velocity is 7m/s (5m/s^2 acceleration*1.4s=7m/s), average velocity is 3.5m/s ((0+7)/2), so distance is 3.5m/s (average) * 1.4s (time of flight): ~5m. This is correct.

So in 1s, A falls 5m. In 1.4s, C falls 5m. So C has a flight time of 1.4 times that of A. Or, A has a flight time of 0.7 (1/1.4) times that of C.

The best way to do these is to practice a hundred of them. Make up your own problems (different initial velocities, different gravities, different heights, find time of flight).

 

[watchntv]

Horizontal velocity is irrelevant to time of flight. The only thing that matters is initial vertical velocity and gravity. In this case, since you make no mention of the vertical velocity, we should assume they both start from vertical rest. We can also assume they start from the same height. This means object A is accelerating downwards at 10m/s^2, and object C is accelerating downwards at 5m/s^2.

Now we have our picture, and we need to figure out the numbers. I will start by saying that finding time of flight from distance is probably the hardest of the kinematics problems to work through, as it eliminates several of your shortcuts. The quickest way I've found is to just do a little guess and check, then round.

For instance, pretend both objects start at vertical position of 5meters. This means object A has a time of flight of 1 second (0m/s initial, 10m/s final, average velocity 5m/s, time of flight 1s, 5m/s*1s=5m). Now do some guess and check to compare object C: at 2s flight time (your guess), initial velocity is 0m/s, final velocity is 10m/s, average velocity is 5m/s, time of flight is 2s, so 5m/s*2s=10m. So object C falls 10m in 2s -- this is too much, so it must be less than that. Let's try 1.4 (the square root of 2, a very common number in physics problems): initial velocity is 0, final velocity is 7m/s (5m/s^2 acceleration*1.4s=7m/s), average velocity is 3.5m/s ((0+7)/2), so distance is 3.5m/s (average) * 1.4s (time of flight): ~5m. This is correct.

So in 1s, A falls 5m. In 1.4s, C falls 5m. So C has a flight time of 1.4 times that of A. Or, A has a flight time of 0.7 (1/1.4) times that of C.

The best way to do these is to practice a hundred of them. Make up your own problems (different initial velocities, different gravities, different heights, find time of flight).

thanks for your time in writing that😀