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\(^2\))
Show Hint
- 1 s
- 2 s
- 3 s
- 4 s
The Correct Option is B
Solution and Explanation
Sum of distances covered equals tower height.
Step 2: Detailed Explanation:
Let time = t
\(h_1 = \frac{1}{2}gt^2\), \(h_2 = ut - \frac{1}{2}gt^2\)
\(h_1 + h_2 = ut = 100\)
\(50t = 100 \rightarrow t = 2\ \mathrm{s}\)
Step 3: Final Answer:
They cross after 2 seconds.
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