Time above ground question involving ratio

AI Thread Summary
The discussion revolves around calculating the ratio of time an athlete spends above half of their maximum jump height (ymax/2) compared to the time taken to reach that height from the ground. Participants clarify that both t1 (time above ymax/2) and t2 (time to reach ymax/2) should be calculated to find the ratio t1:t2. There is confusion about setting the final velocity to zero when calculating time for an object in free fall. One user shares their calculations using the formula for displacement and acceleration due to gravity, leading to a derived maximum height of 1.23 meters. The conversation emphasizes the importance of symmetry in projectile motion to simplify calculations.
Bikengine
Messages
5
Reaction score
0
In the vertical jump, an athlete starts from a crouch and jumps upward to reach as high as possible. Even the best athletes spend little more than 1.00 s in the air (their "hang time"). Treat the athlete as a particle and let ymax be his maximum height above the floor.

To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 moving up to the time it takes him to go from the floor to that height. You may ignore air resistance.

v = u + at
s = ut + 1/2at^2



The wording of this question is really confusing me. So I need to calculate the ratio of the time he is above ymax/2 (t1) : the time it takes him to go from the floor to ymax/2 (t2)? So its just the ratio t1:t2?
 
Physics news on Phys.org
Right.
While it is not necessary, I think it is useful to calculate both t1 and t2, and calculate the ratio based on those two values.
 
mfb said:
Right.
While it is not necessary, I think it is useful to calculate both t1 and t2, and calculate the ratio based on those two values.


Ok I think I've calculated t1, but I don't think it is correct.. if you are doing a calculation for an object being thrown in the air, until it reaches the point where it is coming back down can you set v=0?
 
Bikengine said:
Ok I think I've calculated t1, but I don't think it is correct..
It would be interesting to see your work.
if you are doing a calculation for an object being thrown in the air, until it reaches the point where it is coming back down can you set v=0?
At both points, the object is moving...

If moving up and down together takes 1 second, how long does it take to go up?
 
mfb said:
It would be interesting to see your work.
At both points, the object is moving...

If moving up and down together takes 1 second, how long does it take to go up?

ok so here is what I have done:

s = vt - 1/2at^2 -- I have never used this formula before, found it off google

s = ymax
a= -9.81 m/s^2
t = 0.5 (because that is half the total time to go up and down and I am just finding the up)
v = 0

ymax = 9.81 x 1/2 x 0.5^2 = 1.23m, so ymax/2 = 0.61m
 
That is a good approach. Now you just need the time from the ground to 0.61m, or from 0.61m to 1.23m. Alternatively, use the symmetry and calculate the time from 1.23m to 0.61m.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top