Two balls are dropped from rest from the same height. One of the balls is dropped 1.0 s after the other. What distance separates the two balls 2.00 s after the second ball is dropped?
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To solve this problem, we need to consider the motion of the balls separately.
Let's assume the time when the first ball is dropped as t = 0.
For the first ball:
Time it takes for the first ball to reach a time of 2.00 s after the second ball is dropped = 2.00 s
Using the kinematic equation for free-fall motion:
d = vt + (1/2)gt^2
Where:
d = distance
v = initial velocity (which is 0 since the ball is dropped from rest)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Plugging in the values:
d1 = 0 + (1/2)(9.8)(2.00)^2
Calculating:
d1 = 0 + (1/2)(9.8)(4.00)
d1 = (1/2)(9.8)(4.00)
d1 = 4.00 m
So, the distance between the two balls 2.00 s after the second ball is dropped is 4.00 meters.
To find the distance that separates the two balls 2.00 s after the second ball is dropped, let's first analyze the motion of each ball separately.
Both balls are dropped from rest at the same height. This means they both have an initial velocity of 0 m/s. We know that the acceleration due to gravity is approximately 9.8 m/s^2, which acts downwards for both balls.
Let's consider the first ball. It has been in free fall for 2.00 s by the time the second ball is dropped. The distance traveled by an object in free fall can be calculated using the formula:
d = 0.5 * a * t^2
where d is the distance traveled, a is the acceleration due to gravity, and t is the time elapsed.
Using the given values, we can calculate the distance traveled by the first ball:
d1 = 0.5 * 9.8 * (2.00)^2
= 0.5 * 9.8 * 4.00
= 19.6 m
Now let's consider the second ball. It has only been in free fall for 2.00 - 1.00 = 1.00 s. Using the same formula as above, we can calculate the distance traveled by the second ball:
d2 = 0.5 * 9.8 * (1.00)^2
= 0.5 * 9.8 * 1.00
= 4.9 m
To find the distance that separates the two balls after 2.00 s, we need to subtract the distance traveled by the second ball from the distance traveled by the first ball:
Separation distance = d1 - d2
= 19.6 - 4.9
= 14.7 m
Therefore, the distance that separates the two balls 2.00 s after the second ball is dropped is 14.7 meters.