You drop a rock from a bridge to the river below. When the rock has
fallen 4 m, you drop a second rock. As the rocks continue their free
fall, their separation?
A. increase B. decrease C. stay the same D. none
Answers
Answer:
c) is the correct answer
Answer:
m}.Δh=∣h
2
−h
1
∣=∣0−4∣=4 m.
The time it took for the first rock to cover 4 m is
t_1=\sqrt{\frac{2h_1}{g}}=\sqrt{\frac{2\cdot4}{9.8}}=0.9\text{ s}.t
1
=
g
2h
1
=
9.8
2⋅4
=0.9 s.
The initial velocity of the second rock is of course 0. Let us calculate the distance the second rock covered after 2 seconds:
H_2=\frac{gt^2}{2}=19.6\text{ m}.H
2
=
2
gt
2
=19.6 m.
For the first rock, it will be 2.9 s. The distance of the first rock after 2.9 seconds is
H_1=\frac{9.8\cdot2.9^2}{2}=41.2\text{ m}.H
1
=
2
9.8⋅2.9
2
=41.2 m.
The separation between the rocks 2 second after the second rock was thrown is
\Delta H=|H_2-H_1|=21.6\text{ m}.ΔH=∣H
2
−H
1
∣=21.6 m.
As we see, the separation increases as both rocks continue to fall.