A startled armadillo leaps upward rising 0.540 m in the first 0.195 s.
(a) What is its initial speed as it leaves the ground?
3.7 m/s
(b) What is its speed at the height of 0.540 m?
1.8 m/s
(c) How much higher does it go?
How would I calculate (c)? should I use the formula vf^2=vi^2+2a(yf-yi)?
(a) In 0.195 s, it will slow down by 9.81*0.195 = 1.913 m/s. The average speed during the interval was 2.769 m/s.
[Vo + (Vo-1.91)]/2 = 2.769
Vo - 0.956 = 2.760
Vo = 3.72 m/s
You got that right! Good show!
(b) Vo^2 - V^2 = 2 g H = 10.60 m^2/s^2
V^2 = 3.238 m^2/s^2
V = 1.80 m/s touche!
(c) It rises an additional distance H' such that
2 g H' = V^2
H' = 0.165 m
Thanks!
To calculate how much higher the armadillo goes, you can use the formula vf^2 = vi^2 + 2a(yf - yi), where vf is the final velocity, vi is the initial velocity, a is the acceleration, yf is the final height, and yi is the initial height.
In this case, the armadillo is going upward, so we need to consider the acceleration due to gravity, which is -9.8 m/s^2. The initial height (yi) is 0 m, and the final height (yf) is 0.540 m. We are given the initial velocity (vi) as 3.7 m/s.
Substituting the known values into the formula, we get:
vf^2 = (3.7 m/s)^2 + 2(-9.8 m/s^2)(0.540 m - 0 m)
Simplifying, we have:
vf^2 = 13.69 m^2/s^2 - 10.632 m^2/s^2
vf^2 = 3.058 m^2/s^2
To find vf, we take the square root of both sides:
vf = √(3.058 m^2/s^2)
vf = 1.75 m/s
Now that we have the final velocity at the maximum height, we can calculate how much higher the armadillo goes. We subtract the initial height (0 m) from the final height (0.540 m):
Higher = yf - yi
Higher = 0.540 m - 0 m
Higher = 0.540 m
Therefore, the armadillo goes 0.540 m higher.