Yes, you can! That is what I did. h = 1/2at^2t = sqrt(2h/g) t = sqrt (2v0^2/2g/g) t = sqrt ( 2v0^2/2g^2) The 2 cancels out. And the rest can be removed using square root. t= v0/g
The reasoning is that on way the way up the velocity gets slowed down until it becomes zero by accelereation at a constant rate of -10m/s^2. This all happens within a certain amount of time. so it goes from max velocity to zero on the way up. On the way down it's basically the opposite where velocity goes from 0 to max because accleration of -10 now speeds in up in that same time frame.
I'm a bit confused about part C. If you use the equation or displacement (yf-y0= v0t +1/2 at^2) and since displacement is 0, if you solve for t, you end up with t= 2v0/g. Why would this answer be wrong?
Sir for part a)iv) could we not have said that student 2 wrongly stated that the time for both rockets to land was the same? If not would you mind explaining why? Thanks.
Hello and thanks for the inquiry. Firstly, both rockets do take the same time to reach the ground. y2-y1 = v1*t+(1/2)*a*t^2. Since y2-y1=0 it follows that t=squareroot(2*-v1/-g). Negatives cancel and t=squareroot(2*v1/g). Notice all factors are the same for rocket X and rocket Y (v1 and g) and therefore time is the same. Secondly, if you look again at part a's instructions, it says to ignore whether or not the students' predictions are correct. The time to reach the ground are their predictions, so you don't want to argue "time" assertions. You only want to focus on their reasonings as being correct or incorrect (the height and kinetic energy assertions). Hope that clears it up. Thanks!
uhm so where do i find the correct answers for part d)?
could you also use t=sqrt(2h/g) and substitute the hmax equation (vo^2/g) into the time equation and get vo/g?
Yes, you can! That is what I did.
h = 1/2at^2t = sqrt(2h/g)
t = sqrt (2v0^2/2g/g)
t = sqrt ( 2v0^2/2g^2)
The 2 cancels out. And the rest can be removed using square root.
t= v0/g
can someone explain why in part c) the upward velpcoty at position 2 and the downward one are equal? I didnt quite get what he meant.
The reasoning is that on way the way up the velocity gets slowed down until it becomes zero by accelereation at a constant rate of -10m/s^2. This all happens within a certain amount of time. so it goes from max velocity to zero on the way up. On the way down it's basically the opposite where velocity goes from 0 to max because accleration of -10 now speeds in up in that same time frame.
I'm a bit confused about part C. If you use the equation or displacement (yf-y0= v0t +1/2 at^2) and since displacement is 0, if you solve for t, you end up with t= 2v0/g. Why would this answer be wrong?
I’m not an expert but my teacher said for some problems the kinematic equations can be interchangeable if all variables are present
Sir for part a)iv) could we not have said that student 2 wrongly stated that the time for both rockets to land was the same? If not would you mind explaining why? Thanks.
Hello and thanks for the inquiry. Firstly, both rockets do take the same time to reach the ground. y2-y1 = v1*t+(1/2)*a*t^2. Since y2-y1=0 it follows that t=squareroot(2*-v1/-g). Negatives cancel and
t=squareroot(2*v1/g). Notice all factors are the same for rocket X and rocket Y (v1 and g) and therefore time is the same. Secondly, if you look again at part a's instructions, it says to ignore whether or not the students' predictions are correct. The time to reach the ground are their predictions, so you don't want to argue "time" assertions. You only want to focus on their reasonings as being correct or incorrect (the height and kinetic energy assertions). Hope that clears it up. Thanks!
@@heinrichphysics1976 Thank you for taking the time to explain that I understand it now. Have a good day!