Part 1: Two balls (Ball 1 and Ball 2) are released from the top of a tower. Ball 2 is thrown 3.14 seconds after Ball 1 is dropped. Ball 2 is thrown downward with a velocity of 3.49 m/s. Determine how far apart Ball 1 and Ball 2 are after 5.02 seconds from when Ball 1 was released. (3 pts.)
To determine how far apart Ball 1 and Ball 2 are after 5.02 seconds, we need to calculate the distance each ball has traveled during that time.
1. Calculate the distance Ball 1 has traveled after 5.02 seconds:
Since Ball 1 was dropped and not thrown, it experiences free fall with an acceleration of approximately 9.8 m/s^2 downwards. We can use the equation for distance traveled during free fall:
d1 = (1/2) * g * t1^2, where d1 is the distance traveled by Ball 1, g is the acceleration due to gravity, and t1 is the time Ball 1 has been falling.
Plugging in the values:
d1 = (1/2) * 9.8 m/s^2 * (5.02 s)^2
d1 = 122.98 m (rounded to two decimal places)
2. Calculate the distance Ball 2 has traveled after 5.02 seconds:
Since Ball 2 was thrown downward, its motion can be modeled using the equation:
d2 = v2 * t2, where d2 is the distance traveled by Ball 2, v2 is the velocity of Ball 2, and t2 is the time Ball 2 has been traveling.
Plugging in the values:
d2 = 3.49 m/s * (5.02 s - 3.14 s)
d2 = 6.71 m (rounded to two decimal places)
3. Determine the distance between Ball 1 and Ball 2 after 5.02 seconds:
To find the distance between the balls, we subtract the distance traveled by Ball 2 from the distance traveled by Ball 1:
Distance = d1 - d2
Distance = 122.98 m - 6.71 m
Distance = 116.27 m (rounded to two decimal places)
Therefore, after 5.02 seconds from when Ball 1 was released, Ball 1 and Ball 2 will be approximately 116.27 meters apart.
To determine how far apart Ball 1 and Ball 2 are after 5.02 seconds from when Ball 1 was released, we can calculate the distance each ball has traveled in that time frame and then find the difference between their distances.
First, let's find out how far Ball 1 has fallen after 5.02 seconds. We can use the equation for free-fall distance:
d1 = 1/2 * g * t1^2
where:
d1 = distance fallen by Ball 1
g = acceleration due to gravity (approximately 9.8 m/s^2)
t1 = time Ball 1 has been falling (5.02 seconds)
Plugging in the values, we get:
d1 = 1/2 * 9.8 * (5.02)^2
d1 ≈ 124.01 meters
Next, let's calculate how far Ball 2 has fallen in the 5.02-second time frame. Since Ball 2 was thrown downward, we can use the equation for distance traveled with constant acceleration:
d2 = v0 * t2 + 1/2 * a * t2^2
where:
d2 = distance fallen by Ball 2
v0 = initial velocity of Ball 2 (3.49 m/s, downward)
t2 = time Ball 2 has been falling (5.02 - 3.14 seconds, as Ball 2 was released 3.14 seconds after Ball 1 was dropped)
a = acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the values, we get:
d2 = 3.49 * (5.02 - 3.14) + 1/2 * 9.8 * (5.02 - 3.14)^2
d2 ≈ 11.78 meters
Now, we can find the distance between Ball 1 and Ball 2 after 5.02 seconds by subtracting the distance fallen by Ball 2 from the distance fallen by Ball 1:
Distance = d1 - d2
Distance ≈ 124.01 - 11.78
Distance ≈ 112.23 meters
Therefore, Ball 1 and Ball 2 are approximately 112.23 meters apart after 5.02 seconds from when Ball 1 was released.