Question

Drops of water are released from a 9 m high building at regular intervals. Fourth drop starts to fall as soon as first drop reaches ground. What are the distances between second and third drops?

Answer 1

Time taken by ball to reach the ground is $t=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\times 9}{9.8}}=1.35$ s.
So the time interval by which the ball were released is $\frac{1.35}{4-1}=0.45$ s.
So the time taken by third ball is $2\times 0.45=0.9$ s, hence distance travelled by third ball is $H_3=\frac{1}{2}g{t}^2=\frac{1}{2}\times 9.8\times {\left(0.9\right)}^2=3.97$ m.
Time travelled by second ball is 0.45 s and hence the distance travelled is $H_2=\frac{1}{2}\times 9.8\times {\left(0.45\right)}^2=0.99$ m.
So distance between two ball is 3.97-0.99=2.98 m.