A juggler throws a ball straight up into the air and it takes 2.80 seconds to reach its peak. How many seconds will it take after that to fall back into his hand? Assume he throws and catches the ball at the same height.
Tf = Tr = 2.80 s. = Fall time.
To find out how many seconds it will take for the ball to fall back into the juggler's hand, we need to consider the total time it takes for the ball to go up and then come back down.
We know that it takes 2.80 seconds for the ball to reach its peak. However, this time only accounts for one half of the total time because the ball still needs to come back down.
The total time can be calculated by multiplying the time it took for the ball to reach its peak by 2. This is because the ball takes the same amount of time to come down from its peak as it did to reach it.
So, to find the total time, we multiply 2.80 seconds by 2:
2.80 s * 2 = 5.60 seconds
Therefore, it will take 5.60 seconds for the ball to fall back into the juggler's hand.