Question

A stone falls from the top of a tower of height \(100\) m and at the same time another stone is projected vertically upwards from the ground with a velocity \(25\) m/s. The two stones meet after

• A. \(2\) s
• B. \(5\) s
• C. \(4\) s
• D. \(3\) s

Hint:

Gravity cancels in the relative motion. t = 100/25.

Answer 1

The Correct Option is C

Step 1: Equal positions:
Take upward as positive, with the ground at $y=0$. Stone A: $y_A=100-\frac12gt^2$. Stone B: $y_B=25t-\frac12gt^2$.

Step 2: Equate:
The $\frac12gt^2$ terms cancel. $100=25t$, so $t=4$.

Step 3: Answer:
4 s. At this time the stones are at height $100-\frac12(10)(16)=20$ m (taking $g=10$), which is a sensible value, below the tower top. Option (C).

Final Answer:
The gt^2 terms cancel, so t = 100/25 = 4 s. \[ \boxed{C} \]