Question

A ball is dropped from the top of a building 100 m high. At the same instant another ball is thrown upwards with a velocity of 40 m/s from the bottom of the building. The two balls will meet after

• A. 3 s
• B. 2 s
• C. 2.5 s
• D. 5 s

Answer 1

Answer: (C)

2.5 s

Answer 2

Let balls meet after t s. The distance travelled by the ball coming down is $\quad\quad$ s$_{1}=\frac{1}{2}gt^{2}$ Distance travelled by the other ball $\quad\quad$ s$_{2}=40t-\frac{1}{2}gt^{2}$ $\because \quad s_{1}+s_{2}$ = 100 m $\therefore\quad \frac{1}{2}gt^{2}+40t-\frac{1}{2}gt^{2}$ = 100 m $\quad\quad t=\frac{100}{40}$ = 2.5 s