A marble with speed 14 cm/s rolls off the edge of a table 75 cm high. How long does it take to drop to the floor? How far, horizontally, from the table edge does the marble strike the floor?
1/2 g t^2 = .75 m ... t^2 = 1.5 m / g
... solve for t
horizontal distance = .14 m/s * t
hf=hi+vi*t+1/2 h t^2 with hf=0, hi=.75, g=9.8m/s^2 then
0=.75 -1/2 g t^2
thus 1/2 g t^2=.75
horizontal distance= horizontal velocity*time
=.14m/s * time
To find the time it takes for the marble to drop to the floor, we can use the equation for time:
t = √(2h/g)
where:
t = time (in seconds)
h = height (in meters)
g = acceleration due to gravity (approximately 9.8 m/s²)
Converting the height of the table from centimeters to meters:
h = 75 cm = 0.75 m
Plugging the values into the equation:
t = √(2 * 0.75 m / 9.8 m/s²)
t = √(0.1531 s²)
t ≈ 0.3914 s
Therefore, it takes approximately 0.3914 seconds for the marble to drop to the floor.
To calculate the horizontal distance from the table edge to where the marble strikes the floor, we can use the equation for horizontal distance:
d = v * t
where:
d = horizontal distance (in meters)
v = horizontal velocity (in meters per second) - assumed to be constant throughout the fall
t = time (in seconds) - already calculated as 0.3914 s
Since the marble rolls off the edge of the table, its initial horizontal velocity is equal to its rolling speed of 14 cm/s or 0.14 m/s.
Plugging the values into the equation:
d = 0.14 m/s * 0.3914 s
d ≈ 0.0547 m
Therefore, the marble strikes the floor approximately 0.0547 meters (or 5.47 cm) from the table edge horizontally.
To find the time it takes for the marble to drop to the floor, we can use the formula for free fall:
\[h = \frac{1}{2}gt^2\]
where:
- \(h\) is the height (75 cm in this case)
- \(g\) is the acceleration due to gravity (approximately 9.8 m/s² or 980 cm/s²)
- \(t\) is the time it takes for the marble to fall
First, let's convert the height from centimeters to meters:
\[75 \text{ cm} = 0.75 \text{ m}\]
Now we can plug in the values into the formula:
\[0.75 \text{ m} = \frac{1}{2} \cdot 980 \text{ cm/s}^2 \cdot t^2\]
To solve for \(t\), we can rearrange the equation:
\[t^2 = \frac{2 \cdot 0.75 \text{ m}}{980 \text{ cm/s}^2}\]
\[t^2 = \frac{1.5 \text{ m}}{980 \text{ cm/s}^2}\]
\[t^2 = 0.00153 \text{ s}^2\]
\[t \approx 0.039 \text{ s}\]
Therefore, it takes approximately 0.039 seconds for the marble to drop to the floor.
To find the horizontal distance the marble strikes from the table edge, we can use the formula for horizontal distance:
\[d = v \cdot t\]
where:
- \(d\) is the horizontal distance
- \(v\) is the horizontal velocity (which is constant in this case)
- \(t\) is the time we just found (0.039 s)
The horizontal velocity \(v\) can be found using the initial speed of the marble. Since the marble is rolling off the edge of the table, the horizontal component of its velocity remains constant. Therefore, the horizontal velocity is equal to the initial speed of the marble, which is 14 cm/s.
Now we can plug in the values into the formula:
\[d = 14 \text{ cm/s} \cdot 0.039 \text{ s}\]
\[d \approx 0.546 \text{ cm}\]
Therefore, the marble strikes the floor approximately 0.546 cm horizontally from the table edge.