marble with speed 35 cm/s rolls off the edge of a table 75 cm high. how long does it take to drop to the floor?
We can use the formula:
h = 0.5 * g * t^2
where h is the distance fallen (75 cm in this case), g is the acceleration due to gravity (9.8 m/s^2 or 980 cm/s^2), and t is the time taken to fall.
Rearranging the formula, we get:
t = square root(2h / g)
Substituting the values, we get:
t = square root(2 * 75 / 980) = square root(0.153)
Therefore, the time taken for the marble to fall to the floor is:
t = 0.391 seconds (rounded to three decimal places)
To calculate the time it takes for the marble to drop to the floor, we can use the equations of motion. The height of the table (h) and the initial vertical velocity (u) of the marble are given as 75 cm and 0 cm/s, respectively. The acceleration due to gravity (g) is approximately 9.8 m/s².
First, let's convert the given values to meters:
height (h) = 75 cm = 0.75 m
initial vertical velocity (u) = 0 cm/s = 0 m/s
acceleration due to gravity (g) = 9.8 m/s²
We can use the following equation of motion:
h = ut + (1/2)gt²
Substituting the known values into the equation:
0.75 = 0t + (1/2) * 9.8 * t²
Simplifying the equation:
0.75 = 4.9t²
Rearranging the equation:
4.9t² = 0.75
Dividing both sides by 4.9:
t² = 0.75 / 4.9
t² ≈ 0.153
Taking the square root of both sides:
t ≈ √0.153
t ≈ 0.391 s
Therefore, it takes approximately 0.391 seconds for the marble to drop to the floor.