A child stands on a balcony directly above a friend 5.00
m below. Each throws a ball toward the other with the
same initial speed of 10.0 m/s. If each catches a ball at
the same instant, which child threw a ball first? How
much later was the second ball thrown?
Well, that's quite a toss-up! But I'm here to clown around and give you an answer. Since both children throw the balls at the same initial speed of 10.0 m/s, and they catch the balls at the same instant, it means that the time it takes for each ball to reach the other child is the same.
From the given information, we can calculate the time it takes for the ball to reach the friend below. Using the formula:
time = distance / speed
The time it takes for the ball to reach the friend is:
time = 5.00 m / 10.0 m/s = 0.50 s
So, both children throw the ball at the same time because they catch it simultaneously. Therefore, the concept of who threw the ball first doesn't really apply in this case.
Sorry to burst your bubble, but both balls were thrown at the same exact time!
To determine which child threw the ball first and how much later the second ball was thrown, we can use the concepts of projectile motion and calculating the time of flight for each ball.
The time of flight for a projectile can be calculated using the equation:
๐ก = (2๐)/๐ฃ
Where:
๐ก is the time of flight,
๐ is the vertical distance (in this case, the height of the balcony),
๐ฃ is the initial vertical velocity (which is zero when released).
For the child on the balcony, the time of flight can be calculated as:
๐กโ = (2๐)/๐ฃ = (2 * 5.00 m)/(10.0 m/s) = 1.00 s
For the child below, the time of flight can be calculated as:
๐กโ = (2๐)/๐ฃ = (2 * 5.00 m)/(10.0 m/s) = 1.00 s
Since both children threw the balls with the same initial speed, and the time of flight for each ball is the same, we can conclude that both balls were thrown at the same instant.
Therefore, both children threw the balls at the same time, and there was no time difference between their throws.