Near the surface of the Earth, a startled armadillo leaps vertically upward at time t = 0, at time t = 0.5 s, it is a height
of 0.98 m above the ground. At what time does it land back on the ground?
To find the time the armadillo lands back on the ground, we need to determine how long it takes for the armadillo to complete its upward and downward journey.
Let's assume the armadillo follows a parabolic trajectory. We can use the equation of motion for an object in free fall, taking into account that the initial velocity is zero since the armadillo starts from rest.
The equation to calculate the height of an object in free fall is:
h = (1/2)gt^2
Where:
h = height of the object
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Since the initial velocity is zero, the maximum height reached by the armadillo will be at the halfway point in time.
Given that the armadillo reaches a height of 0.98 m at t = 0.5 s, we can determine the maximum height reached.
0.98 m = (1/2)(9.8 m/s^2)(0.5 s)^2
Simplifying the equation:
0.98 m = 1.225 m
The armadillo reaches a maximum height of 1.225 m.
Now, we can find the time it takes for the armadillo to travel from the maximum height to the ground, using the same equation:
0 = (1/2)(9.8 m/s^2)t^2
Rearranging the equation:
0 = 4.9t^2
Simplifying the equation:
t^2 = 0
This indicates that t = 0 seconds. Thus, the armadillo lands back on the ground at t = 0 seconds, or instantaneously.
Therefore, the armadillo lands back on the ground as soon as it reaches its maximum height.