When do the two balls pass each other in the annoying juggler problem?

[DB]

a juggler performs in a room whose ceiling is 3 m above the level of his hands. he throws a ball vertically upward so that it JUST reaches the ceiling. he then throws a second ball upward with the same inital velocity, at the instant the first ball is at the ceiling. How long after the second ball is thrown do the two balls pass each other?

okay, so I've gotten that the intial velocity of the balls is 7.75 m/s and that the time required for a ball to hit the ceiling is 0.775 seconds. the answer sheet tells me this is right. so i thought that, considering the situation, the balls would pass each other halfway through their trips, at 1.5 m. so i solved the quadratic that gave me how long it would take the second ball to reach 1.5 m and i got 0.175 seconds. this is wrong. how come? if they don't pass each other halfway through their trips then when?

thanks in advance

 

[Aneleh]

You're trying to solve for when they pass each other, you can't just guess (well you can but...) at 1.5m they pass. If you imagine the paths followed by the two balls and you want to find the point where they have the same height, use your quadratic equations for both balls by equating the height, then solve for the time.

 

[quicknote]

I would first like to commend you on your efforts to solve the problem and your attention to detail in calculating the initial velocity and time required for a ball to reach the ceiling. However, it seems like there may be a misunderstanding about the concept of projectile motion.

In this scenario, both balls are thrown with the same initial velocity and in the same direction, but at different times. This means that the second ball will not necessarily reach the halfway point at the same time as the first ball. In fact, the second ball will reach the ceiling at a different time than the first ball, as it was thrown later.

To calculate the time at which the two balls pass each other, we need to consider the vertical displacement of each ball. The first ball will have a displacement of 3 meters (the height of the ceiling) and the second ball will have a displacement of 1.5 meters (halfway point between the initial position and the ceiling). We can use the equation for displacement in projectile motion (d = v0t + 1/2at^2) to solve for the time when each ball reaches their respective displacements.

When we set the displacement of the first ball to 3 meters, we get a time of 0.775 seconds, which is the same as the time calculated for the ball to reach the ceiling. However, when we set the displacement of the second ball to 1.5 meters, we get a time of approximately 0.329 seconds. This means that the two balls will pass each other after 0.329 seconds, not at the halfway point of their trips.

In summary, it is important to consider the individual displacement and time for each ball in projectile motion problems, rather than assuming they will pass each other at the halfway point. I hope this explanation helps to clarify the solution to the annoying juggler problem. Keep up the good work in your scientific pursuits!