An object falls a distance h from rest. If it travels 0.45h in the last 1.00 s, find the time and the height of its fall.
h=v_0+at^2
h=0+(-9.8)t^2
h=96.4(t)
.45h=96.4(1)
h=96.1/0.45
h=213.422ft
use this equation:
(0.45)h=(1/2)*g*(2t-1)
To find the time and height of the object's fall, we can use the equations of motion.
The given information tells us that in the last 1.00 second, the object falls a distance of 0.45h. Let's denote this distance as d.
Now, we know that the distance traveled by a falling object can be calculated using the equation:
d = ut + (1/2)at^2
Where:
- d is the distance traveled
- u is the initial velocity (which is 0 in this case because the object is at rest)
- t is the time taken
- a is the acceleration due to gravity (which is approximately 9.8 m/s^2)
In this case, we can rearrange the equation to solve for time:
t = sqrt(2d/a)
Substituting the given value of d = 0.45h into the equation, we can now solve for time.
t = sqrt(2 * 0.45h / 9.8)
To find the height of the fall (h), we will multiply the time (t) by 2 since the total fall time is twice the time taken for the last 1.00 second.
h = 2 * t
Now, let's substitute the value of t into the equation to find the height of the fall.
h = 2 * sqrt(2 * 0.45h / 9.8)
Simplifying this equation will give us the value of h.
This is a nonlinear equation, so we'll need to use approximation methods or numerical methods to find the value of h.