Question

A startled armadillo leaps upward, rising 0.537 m in the first 0.220 s. (a) what is its initial speed as it leaves the ground? (b) what is its speed at the height of 0.537 m? (c) how much higher does it go? use g=9.81 m/s2.

Answer 1

Let u = initial vertical velocity.

Assume that
g = 9.81 m/s²,
Wind resistance is ignored.

When t = 0.220 s, the height is h = 0.537 m. Therefore
0.537 m = (u m/s)*(0.220 s) - (1/2)*(9.81 m/s²)*(0.220 s)²
0.537 = 0.22u - 0.2372
u = 3.519 m/s

The upward velocity after 0.220 s is
v = 3.519 - 9.81*0.22 = 1.363 m/s

At maximum height, the upward velocity is zero. The maximum height, H, is given by
(3.519 m/s)² - 2*(9.81 m/s²)*(H m) = 0
12.3834 - 19.6H = 0
H = 0.632 m
It goes higher by 0.632 - 0.537 = 0.095 m

Answers:
(a) The initial speed is 3.519 m/s.
(b) The speed at  0.537 m height is 1.363 m/s.
(c) It goes higher by 0.095 m.