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.)
no
To determine how far apart Ball 1 and Ball 2 are after 5.02 seconds, we need to consider the motion of both balls separately and then find the difference in their positions.
First, let's analyze the motion of Ball 1. Ball 1 is simply dropped from the top of the tower, so its initial velocity is 0 m/s. 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 (approximately 9.8 m/s^2), and t1 is the time Ball 1 has been falling (5.02 seconds in this case).
Now, let's analyze the motion of Ball 2. Ball 2 is thrown downward with a velocity of 3.49 m/s. We can use the equation for distance traveled with constant velocity:
d2 = v2 * t2
Where d2 is the distance traveled by Ball 2, v2 is the velocity of Ball 2 (-3.49 m/s since it is thrown downward), and t2 is the time Ball 2 has been thrown (5.02 seconds - 3.14 seconds = 1.88 seconds).
Finally, to find the distance between Ball 1 and Ball 2, we subtract the distance traveled by Ball 2 from the distance traveled by Ball 1:
Distance = d1 - d2
By plugging in the values, we can calculate the distance between the two balls after 5.02 seconds.