Question

A ball is dropped from top of a tower of 100 m height. Simultaneously another ball was thrown upward from bottom of the tower with a speed of 50 m/s. They will cross each other after (g = 10 m/s\(^2\))

• A. 1 s
• B. 2 s
• C. 3 s
• D. 4 s

Hint:

The acceleration terms cancel, making meeting time independent of g.

Answer 1

The Correct Option is B

To solve this problem, let's analyze the motion of both balls using the equations of motion under gravity.

1. The first ball is dropped from the top of a tower of 100 m height. Its initial velocity (\(u_1\)) is 0 m/s, since it is dropped. It is subjected to a gravitational acceleration (\(g\)) of 10 m/s² downward.
2. The second ball is thrown upward from the bottom with an initial speed (\(u_2\)) of 50 m/s, and it is subjected to the same gravitational acceleration (\(g\) = -10 m/s², negative because it is in the opposite direction to the motion).
3. We'll use the equation of motion: \(s = ut + \frac{1}{2} a t^2\).
4. For the first ball: The displacement upwards (\(s_1\)) after time \(t\) is given by: 
   \(s_1 = u_1 \cdot t + \frac{1}{2} g t^2 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 = 5t^2\).
5. For the second ball: The displacement upwards (\(s_2\)) is given by: 
   \(s_2 = u_2 \cdot t + \frac{1}{2} (-g) t^2 = 50t - 5t^2\).
6. The total displacement when they meet should equal the height of the tower, i.e.: 
   \(s_1 + s_2 = 100\).
7. Substitute the expressions for \(s_1\) and \(s_2\): 
   \(5t^2 + (50t - 5t^2) = 100\).
8. Simplify: 
   \(50t = 100\).
9. Solving for \(t\) gives: 
   \(t = \frac{100}{50} = 2\) seconds.

Thus, the two balls will cross each other after 2 seconds.