To solve this problem, we need to analyze the motion of both the lift and the ball separately, then find the time when they meet again. Let's break this down step by step.
Understanding the Motion of the Lift
The lift is moving upwards with an initial velocity of 10 m/s and has an upward acceleration of 2 m/s². We can use the equations of motion to describe its position over time.
Position of the Lift
The formula for the position of an object under constant acceleration is:
s = ut + (1/2)at²
- s = displacement
- u = initial velocity (10 m/s)
- a = acceleration (2 m/s²)
- t = time in seconds
Substituting the values, we get:
s_lift = 10t + (1/2)(2)t² = 10t + t²
Analyzing the Ball's Motion
The ball is projected upwards with an initial velocity of 20 m/s relative to the ground. Since the lift is also moving upwards, we need to consider the ball's motion relative to the ground as well.
Position of the Ball
Using the same equation of motion, the position of the ball can be expressed as:
s_ball = ut + (1/2)at²
Here, the initial velocity u is 20 m/s, and the acceleration a is -9.8 m/s² (since gravity acts downwards). Thus, we have:
s_ball = 20t - (1/2)(9.8)t² = 20t - 4.9t²
Finding the Time When the Ball Meets the Lift
To find the time when the ball meets the lift again, we set their positions equal to each other:
10t + t² = 20t - 4.9t²
Rearranging this gives:
0 = 20t - 10t - t² - 4.9t²
0 = 10t - 5.9t²
Factoring out t:
t(10 - 5.9t) = 0
This gives us two solutions: t = 0 (the initial moment) and t = 10/5.9 ≈ 1.69 seconds.
Calculating Displacement at the Meeting Time
Now that we have the time, we can find the displacement of both the lift and the ball at this time.
Displacement of the Lift
Using the time t ≈ 1.69 seconds in the lift's position equation:
s_lift = 10(1.69) + (1/2)(2)(1.69)²
s_lift ≈ 16.9 + 1.69 ≈ 18.59 meters
Displacement of the Ball
Now, substituting the same time into the ball's position equation:
s_ball = 20(1.69) - 4.9(1.69)²
s_ball ≈ 33.8 - 12.78 ≈ 21.02 meters
Distance Travelled by the Ball
To find the distance travelled by the ball up to the instant it meets the lift, we can calculate the distance it travels upwards before it starts descending. The time taken to reach the maximum height can be found using:
v = u + at
Setting the final velocity v to 0 (at the peak):
0 = 20 - 9.8t
This gives us t ≈ 2.04 seconds to reach the maximum height. The distance to the maximum height is:
s_max = 20(2.04) - 4.9(2.04)² ≈ 20.4 - 20.4 = 0 meters
Since the ball meets the lift before reaching its maximum height, we only consider the distance travelled in the time t ≈ 1.69 seconds:
Distance travelled by the ball = 20(1.69) - 4.9(1.69)² ≈ 21.02 meters
Summary of Results
- a) Time when the ball meets the lift: ≈ 1.69 seconds
- b) Displacement of the lift and ball at that time: Lift ≈ 18.59 meters, Ball ≈ 21.02 meters
- c) Distance travelled by the ball up to that instant: ≈ 21.02 meters
This comprehensive breakdown should give you a clear understanding of how to approach problems involving relative motion and the equations of motion. If you have any further questions or need clarification on any part, feel free to ask!
