It takes 0.220 s for a dropped object to pass a window that is 1.14 m tall. From what height above the top of the window was the object released?
Let d be the distance from the drop point to the top of the window.
The time to reach the top is
t = sqrt(2d/g)
The time to reach the bottom is
t' = t + 0.220 = sqrt[2(d+1.14)/g]
= sqrt(2d/g) + 0.22
It's rather messy, but there you have one equation in the single unknown, d. Solve for d.
Let:
d1 = distance from drop to top of window
d2 = distance from drop to bottom of window
t1 = time from ball drop to top of window
t2 = time from ball drop to bottom of window
vi = initial velocity
g = acceleration due to gravity
Knowing the equation (d = vi*t + (a*t^2)/2)
d1 = (g*t1^2)/2
d2 = (g(t1+.22)^2)/2
d2 = d1 + 1.14
SO:
(g(t1+.22)^2)/2 = (g*t1^2)/2 + 1.14
Solve for t1.
Use t1 = sqrt(2*d1/g)
Solve for d1 and you will have the distance from the drop to the top of the window.
To calculate the initial height from which the object was released, we need to use the equation of motion for free-falling objects:
h = 1/2 * g * t^2,
where:
h is the initial height,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time taken for the object to pass the window (0.220 s).
First, we need to calculate the time it takes for the object to fall from the top of the window to the ground. Since the object passes the window in 0.220 s, the total time for the object to reach the ground is twice that,
t_total = 2 * 0.220 s = 0.440 s.
Next, we plug this time value into the equation and solve for the initial height:
h = 1/2 * g * t_total^2.
Substituting the known values,
h = 1/2 * 9.8 m/s^2 * (0.440 s)^2,
h = 0.5 * 9.8 m/s^2 * 0.1936 s^2,
h ≈ 0.9515 m.
Therefore, the object was released from a height of approximately 0.9515 meters above the top of the window.
To find the height above the top of the window from which the object was released, we can use the formula for the height of fall. The formula is:
height = (1/2) * g * t^2
Where:
- height is the height above the top of the window (our unknown)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time it takes for the object to pass the window (0.220 s)
First, let's calculate the height:
height = (1/2) * 9.8 m/s^2 * (0.220 s)^2
height = 0.5 * 9.8 m/s^2 * (0.0484 s^2)
height = 4.9 m/s^2 * 0.0484 s^2
height = 0.23716 m
Therefore, the object was released from a height of approximately 0.23716 meters above the top of the window.