Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t = 0 s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2 s. After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ms⁻¹. [use g = 10 ms⁻²] :

Solution

Physics sir · 4 steps

Use ball A to find when they meet, then give ball B only the 2 s it actually spends in the air.

1. Distance fallen. They meet 100 m above the ground, so each ball has fallen 180 - 100 = 80 m from the top. Take downward as positive.
2. Time for A. Ball A starts from rest at t = 0, so 80 = 1/2(10)t². This gives t² = 16, so they meet at t = 4 s.
3. Time for B. Ball B leaves at t = 2 s, so it is in the air for only 4 - 2 = 2 s. It must cover the same 80 m in that time.

   80 = u(2) + 1/2(10)(2)² = 2u + 20

4. So. 2u = 60, so u = 30 m s⁻¹, which is option (d).

The idea to keep. When two bodies start at different times, fix the meeting moment using the simpler body first. Then use the other body's own flight time, not the clock reading.

Answer from the NTA final answer key. Solution by PrepWiser.