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: Understanding the Concept:
Both stones move under the same acceleration \(g\), so relative to each other they move at constant relative speed. This simplifies the meeting time.

Step 2: Relative motion:
The falling stone starts at rest, and the other starts at \(25\) m/s upward. Gravity affects both equally, so their separation shrinks at a constant rate equal to the upward speed, \(25\) m/s.

Step 3: Compute:
Initial gap is \(100\) m. So
\[ t=\frac{100}{25}=4\ \text{s} \]

Step 4: Check with equations:
Stone 1 falls: \(y_1=100-\tfrac12gt^2\). Stone 2 rises: \(y_2=25t-\tfrac12gt^2\). Setting \(y_1=y_2\) gives \(100=25t\), so \(t=4\) s.

Step 5: Choose:
Option (C).

Final Answer:
The stones meet after 4 s. \[ \boxed{4\ \text{s}} \]