A ball is dropped from the top of a building 100m high. At the same instant another ball is thrown upwards with a velocity of 40 ms-1 from the bottom of the building. The two balls will meet after A. 5 sec B. 2.5 sec C. 2 sec D. 3 sec

To find out when the two balls will meet, we can analyze the motion of each ball separately and then set their positions equal to each other. Let's break this down step by step.

The Scenario

We have two balls: one is dropped from the top of a 100-meter building, and the other is thrown upwards with an initial velocity of 40 m/s from the ground level. The equations of motion will help us determine when they intersect.

Ball Dropped from the Top

The first ball is dropped, meaning it has an initial velocity of 0 m/s. It falls under the influence of gravity, which we will approximate as 9.81 m/s². The equation of motion for this ball can be expressed as:

• Distance fallen = Initial velocity × time + (1/2) × acceleration × time²

Plugging in the values, we get:

• Distance fallen = 0 × t + (1/2) × 9.81 × t²
• Distance fallen = 4.905t²

Since the total height is 100m, the position of the first ball from the top of the building after time t can be expressed as:

• Height of the first ball = 100 - 4.905t²

Ball Thrown Upwards

The second ball is thrown upwards with an initial velocity of 40 m/s. For this ball, the equation of motion is:

• Distance = Initial velocity × time - (1/2) × acceleration × time²

Substituting the values yields:

• Distance = 40t - (1/2) × 9.81 × t²
• Distance = 40t - 4.905t²

When considering the position from the bottom of the building, the height of the second ball as a function of time is:

• Height of the second ball = 40t - 4.905t²

Finding the Meeting Point

To find when the two balls meet, we need to set their positions equal to each other:

• 100 - 4.905t² = 40t - 4.905t²

Notice that the terms with 4.905t² cancel out, simplifying our equation to:

• 100 = 40t

Now, solving for t gives us:

• t = 100 / 40
• t = 2.5 seconds

Conclusion

The two balls will meet after 2.5 seconds. Therefore, the correct answer is B: 2.5 seconds. This problem illustrates how we can use the equations of motion to analyze the dynamics of objects in free fall and those projected upwards, helping us understand the interactions of their paths in a clear manner.