Answer:
3.41 s
114 m
Explanation:
The object is falling in free fall, accelerated by the surface gravity of Earth. We can use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
We set up a frame of reference with the origin at the point the object was released and the X axis pointing down. Then X0 = 0. Since the problem doesnt mention an initial speed we assume V0 = 0.
It travels 0.5h in the last 1 second of the fall. This means it also traveled in the rest of the time of the fall. t = t1 is the moment when it traveled 0.5*h.
0.5*h = 1/2 * a * t1^2
h = a * t1^2
It travels 0.5*h in 1 second.
h = X(t1 + 1) = 1/2 * a * (t1+1)^2
Equating both equations:
a * t1^2 = 1/2 * a * (t1+1)^2
We simplify a and expand the square
t1^2 = 1/2 * (t1^2 + 2*t1 + 1)
t1^2 - 1/2 * t1^2 - t1 - 1/2 = 0
1/2 * t1^2 - t1 - 1/2 = 0
Solving electronically:
t1 = 2.41 s
total time = t1 + 1 = 3.41.
Now
h = a * t1^2
h = 9.81 * 3.41^2 = 114 m
