Chapter 02, Problem 044 Partially correct answer. Your answer is partially correct. Try again. A startled armadillo leaps upward, rising 0.565 m in the first 0.199 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.565 m? (c) How much higher does it go? Use g=9.81 m/s2.

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

Answer sample (c); // would be best because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest;