A stone is dropped from the roof of a high building. A second stone is dropped 1.50 s later. How far apart are the stones when the second one has reached a speed of 12.0 m/s?
The correct answer and explanation is :
To solve this problem, we analyze the motion of both stones using kinematic equations. The motion of both stones follows free fall, meaning their acceleration is due to gravity, ( g = 9.81 \, \text{m/s}^2 ).
Step 1: Determine the Time for the Second Stone
We use the velocity equation for free fall:
[
v = g t
]
Solving for ( t ) when the second stone reaches a speed of 12.0 m/s:
[
t = \frac{v}{g} = \frac{12.0}{9.81} = 1.22 \text{ s}
]
So, the second stone has been falling for 1.22 seconds.
Step 2: Find the Distance Fallen by Each Stone
The displacement of a freely falling object is given by:
[
y = \frac{1}{2} g t^2
]
Distance Fallen by the Second Stone:
[
y_2 = \frac{1}{2} (9.81) (1.22)^2
]
[
y_2 = 7.30 \text{ m}
]
Distance Fallen by the First Stone:
Since the first stone was dropped 1.50 s earlier, its total fall time is:
[
t_1 = 1.22 + 1.50 = 2.72 \text{ s}
]
Using the displacement equation:
[
y_1 = \frac{1}{2} (9.81) (2.72)^2
]
[
y_1 = 36.3 \text{ m}
]
Step 3: Find the Distance Between the Two Stones
[
\Delta y = y_1 – y_2 = 36.3 – 7.30 = 29.0 \text{ m}
]
So, when the second stone reaches a speed of 12.0 m/s, the two stones are 29.0 meters apart.
Here is an illustration showing the two stones being dropped from a tall building, with the first stone farther down and the second one higher up but accelerating downward. The vertical separation of 29.0 meters is marked between them.