A startled armadillo leaps upward, rising 0.515 m in the first 0.203 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.515 m? (c) How much higher does it go?

hf=hi+vi*t-4.8t^2 you know hi=0, hf, t solve for vi

at the top, vf=0
vf^2=vi^2+2gh where g=-9.8 solve for h.

h = Vi t - 4.9 t^2

.515 = Vi (.203) -4.9 (.041)
.203 Vi = .717
Vi = 3.53 m/s

(b) v = Vi - 9.8 t

(c) 0 = Vi - 9.8 t solve for t at top
then
h = Vi t - 4.9 t^2

To answer these questions, we can use the equations of motion. There are three key equations we can utilize:

1. The first equation relates displacement, initial velocity, time, and acceleration:
𝑑 = 𝑢𝑡 + 0.5𝑎𝑡^2
2. The second equation relates final velocity, initial velocity, acceleration, and displacement:
𝑣^2 = 𝑢^2 + 2𝑎𝑑
3. The third equation relates final velocity, initial velocity, acceleration, and time:
𝑣 = 𝑢 + 𝑎𝑡

(a) To find the initial speed (𝑢), we need to use the first equation with the given values:
𝑑 = 0.515 m, 𝑡 = 0.203 s, and assuming 𝑎 (acceleration) to be the acceleration due to gravity (-9.8 m/s^2).
0.515 = 𝑢(0.203) + 0.5(-9.8)(0.203)^2
Solve this equation to find 𝑢.

(b) To find the speed at the height of 0.515 m, we can use the second equation. Since the armadillo is at the highest point of its motion, its final velocity (𝑣) will be zero.
𝑣^2 = 𝑢^2 + 2𝑎𝑑
Substitute 𝑢 (initial velocity), 𝑎 (acceleration due to gravity), and 𝑑 (displacement) into this equation.
0 = 𝑢^2 + 2(-9.8)(0.515)
Solve this equation to find 𝑢.

(c) To find how much higher the armadillo goes, we need to determine the maximum height it reaches. We can use the first equation again, but this time we know the final velocity is zero.
0 = 𝑢(𝑡) + 0.5(-9.8)(𝑡)^2
This equation gives you the time (𝑡) it takes for the armadillo to rise to its maximum height. Use this value to determine the displacement (maximum height) from the equation in part (a).
Finally, subtract the initial displacement (0.515 m) from the maximum height to find how much higher the armadillo goes.