A startled armadillo leaps upward rising 0.547 m in the first 0.202 s.
(a) What is its initial speed as it leaves the ground?
_____m/s
(b) What is its speed at the height of 0.547 m?
_____m/s
(c) How much higher does it go?
_____m
Use distance vs time:
d=vi*t - 1/2 g t^2 solve for vi.
THen, for the second
d=vi*t -1/2 g t^2 solve for t, then
v=vi*t solve for v.
Finally, knowing vi, solve for the time to the top, where vf is zero.
Vf=vi*t -g t solve for t
height=vi*t - 1/2 g t^2 solve for height.
To find the answers to these questions, we can use the distance vs. time equations of motion and solve for the required variables step by step.
(a) To find the initial speed as the armadillo leaves the ground, we can use the equation:
d = vi * t - (1/2) * g * t^2
Given that the armadillo rises 0.547 m in 0.202 s, we can plug in these values into the equation:
0.547 = vi * 0.202 - (1/2) * 9.8 * (0.202)^2
To find vi, rearrange the equation:
vi = (0.547 + (1/2) * 9.8 * (0.202)^2) / 0.202
(b) To find the speed at the height of 0.547 m, we can use the equation:
d = vi * t - (1/2) * g * t^2
Since the armadillo has reached the height at which it started, the distance d would be 0.547 m. We can rearrange the equation to solve for t:
t = (vi ± √(vi^2 - 4 * (-0.5 * g) * (-0.547))) / (2 * (-0.5 * g))
By substituting the obtained value of vi from the previous step, we can find t.
Finally, we can find the speed at that height using the equation:
v = vi * t
(c) To find how much higher the armadillo goes, we need to determine the time it takes to reach the top of its trajectory. At the top, the final velocity vf is zero. We can use the equation:
vf = vi * t - g * t
Rearranging the equation, we get:
t = vf / (vi - g * t)
By substituting the values of vf (which is zero) and the obtained value of vi from the first part, we can solve for t. Once we have t, we can find the maximum height (h) using the equation:
h = vi * t - (1/2) * g * t^2
By substituting the values into the equation, we can find the answer.
Remember to use the given acceleration due to gravity (g) of 9.8 m/s^2 in the calculations.