An object is dropped from a height H. During the final second of its fall, it traverses a distance of 57.8 m. What was H?

h positive up

h(t) = Hi + Vi t -4.9 t^2
0 = 57.8 + Vi*1-4.9*1^2
Vi = -52.9 m/s at 57.8 m
so how far did it fall to get to that speed and height?
v = -gt
-52.9 = -9.8 t
t = 5.40 seconds
how far did it fall in those first 5.4 seconds? 4.9 * 5.4^2
57.8 = Hi - 4.9 * 5.4^2
Hi = 143 + 58 = 201 m

To determine the height (H) from which the object was dropped, we can use the equation of motion for free fall:

h = 0.5 * g * t^2

where:
h is the distance fallen (57.8 m),
g is the acceleration due to gravity (9.8 m/s^2),
t is the time taken to fall (1 s).

To find H, we need to rearrange the equation:

H = h + 0.5 * g * t^2

Substituting the known values into the equation:

H = 57.8 + 0.5 * 9.8 * 1^2

Simplifying the equation by performing the calculations:

H = 57.8 + 0.5 * 9.8 * 1
H = 57.8 + 4.9
H = 62.7

Therefore, the object was dropped from a height of 62.7 meters.

To find the height (H) from which the object was dropped, we can use the equations of motion. In this case, we'll use the equation for the distance covered during free fall:

d = (1/2) * g * t^2

Where:
- d is the distance covered
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken

Given that the object traveled a distance of 57.8 m in the final second of its fall, we have:

d = 57.8 m
t = 1 s
g = 9.8 m/s^2

Plugging these values into the equation, we can solve for H:

57.8 m = (1/2) * 9.8 m/s^2 * (1 s)^2

First, we square the time:

57.8 m = (1/2) * 9.8 m/s^2 * 1 s

Then multiply the acceleration due to gravity with the squared time:

57.8 m = 4.9 m/s^2 * 1 s

48.9 m = 4.9 m/s^2 * s

Divide both sides of the equation by 4.9 m/s^2:

s = 10

Therefore, the height (H) from which the object was dropped is 10 meters.

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