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
Both balls have the same acceleration $g$, so relative to each other they move at a constant $20$ m/s. They close the $60$ m gap in
$$t = \frac{60}{20} = 3\ \text{s}$$
The dropped ball has fallen $\dfrac{1}{2}(10)(3)^2 = 45$ m in this time, so they meet at $60 - 45 = 15$ m above the ground.
(Check: the other ball is at $20(3) - 5(3)^2 = 15$ m.)
Solution by Sreeraj P, M.Sc Physics