a body travels half of its total path in the last second of its free fall from rest.what is the duration of fall .
mistake)) it has initial velocity
To determine the duration of fall, we need to analyze the motion of the body during free fall.
Let's denote the total duration of fall as "t" seconds.
During the last second of free fall, the body covers half of its total path. This means that in the remaining time (t-1) seconds, the body covers the other half of its total path.
Now, let's use the equations of motion for free fall:
The distance traveled (s) by a body in free fall can be calculated using the equation:
s = (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Using this equation, we can calculate the distance covered in the last second.
For the last second of fall:
s_last_second = (1/2) * g * (t-1)^2
For the remaining time (t-1) seconds:
s_remaining = (1/2) * g * (t-1)(t-1) = (1/2) * g * (t-1)^2
Since we are given that the distance traveled in the last second is half of the total distance:
s_last_second = (1/2) * s_total_distance
(1/2) * g * (t-1)^2 = (1/2) * g * t^2
Now, we can solve for t:
(1/2) * g * (t-1)^2 = (1/2) * g * t^2
(t-1)^2 = t^2
t^2 - 2t + 1 = t^2
-2t + 1 = 0
2t = 1
t = 1/2
Therefore, the duration of the fall is 0.5 seconds or half a second.
0.5d = Vo*t + 0.5gt^2,
0.5d = 0 + 0.5*9.8(1)^2,
0.5d = 4.9,
d = 9.8m.
d = Vo*t + 0.5gt^2 = 9.8m,
0 + 0.5*9.8t^2 = 9.8,
4.9t^2 = 9.8,
t^2 = 2,
t = 1.414s.