To tackle this problem, we need to analyze the motion of both balls separately and determine the velocity required for the second ball to meet the first ball after it has been thrown. Let’s break it down step by step.
Understanding the Motion of the First Ball
The first ball is thrown vertically downward with an initial speed of 5 m/s. We need to calculate its position after a total of 4 seconds since the second ball is thrown 2 seconds later.
- Initial Velocity (u1): 5 m/s
- Acceleration (a): 9.81 m/s² (acceleration due to gravity)
- Total Time (t1): 4 s (2 s before the second ball is thrown and 2 s after)
We can use the equation of motion to find the displacement (s1) of the first ball:
s1 = u1 * t1 + 0.5 * a * t1²
Substituting the values, we get:
s1 = 5 * 4 + 0.5 * 9.81 * 4²
s1 = 20 + 0.5 * 9.81 * 16
s1 = 20 + 78.48 = 98.48 m
Motion of the Second Ball
The second ball is thrown 2 seconds after the first one, meaning it will have 2 seconds of free fall to meet the first ball. To ensure both balls meet after 2 seconds of the second ball’s throw, we need to calculate its required initial velocity (u2).
- Time (t2): 2 s
- Acceleration (a): 9.81 m/s²
- Displacement (s2): The second ball also needs to cover the same vertical distance as the first ball, which is 98.48 m.
Using the same equation of motion for the second ball:
s2 = u2 * t2 + 0.5 * a * t2²
Since s2 must equal 98.48 m, we set up the equation:
98.48 = u2 * 2 + 0.5 * 9.81 * 2²
Calculating the second term:
0.5 * 9.81 * 4 = 19.62 m
Now substituting this back into the equation:
98.48 = 2u2 + 19.62
Rearranging gives:
2u2 = 98.48 - 19.62
2u2 = 78.86
u2 = 39.43 m/s
Determining the Closest Answer
The closest option to our calculated initial velocity of 39.43 m/s is 40 m/s. Therefore, the required velocity of the second ball, so that it meets the first ball after 2 seconds of its throw, is:
40 m/s
This methodical approach allows us to use the fundamental equations of motion effectively, leading us to the correct answer while reinforcing principles of kinematics.
