Q 11-02-043JEE MainMedium
A ball is dropped from the top of a building $60$ m tall. At the same instant, another ball is thrown vertically upward from the ground directly below with a speed of $20$ m/s. Find the height above the ground, in metres, at which they meet. ($g = 10\ \text{m/s}^2$)
Numerical value type. Enter your answer.
Answer: 15
Concept. Both balls have the same acceleration, so relative to each other they move at a constant $20$ m/s.
Step 1. Meeting time. $t = \dfrac{60}{20} = 3$ s.
Step 2. Height. The dropped ball has fallen $\dfrac12(10)(9) = 45$ m, so they meet at $60 - 45 = 15$ m.
Answer. $15$.
Solution by Sreeraj P, M.Sc Physics