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Question

When startled, an armadillo will leap upward. Suppose it rises 0.544 m in the first 0.200 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.544 m? (c) How much higher does it go?

Answer 1

For anything in freefall on Earth (t = seconds):

Acceleration (A) = -9.8 m/s^2
Velocity (V) = (-9.8t + V(initial)) m/s
Position (P) = (-4.9t^2 + V(initial)t + P(initial)) m

P=.544
t=.2

Use position equation to determine V(initial):
.544 = -4.9(.2^2) + .2V(initial)
.544 = -.196 + .2V(initial)
.2V (initial) = .740 m/s
V(initial) = 3.7 m/s (part a answer)

at .544 m, t = .2

Use that t in V equation, to determine V at that time:
V = -9.8t + V(initial) = -1.96 + 3.7 = 1.74 m/s (part b answer)

Maximum height is P when V = 0. Use 0 for V, in V equation, and solve for t:
0 = -9.8t + 3.7
-3.7 = -9.8t
t=9.8/3.7 = .378 seconds

Now use that t in P equation, to solve for P(max)
P= -4.9(.378^2) + .378 x 3.7 = .698 m (part c answer)

Answer 2

apply s = ut-1/2gt^2

put s = 0.544
t= 0.2
solve for u, the initial velocity

(b) calculate the time required to reach height 0.544 with initial velocity calculated as above using the same formula

then apply v = u - gt for velocity

(c) total height = u^2/ 2g

Answer 3

a) 1/2 M Vo^2 = 1/2 M V^2-MGH
Vo= sqrt (V^2-GH)
where V= .544/.2 and G=9.8 and H=.544

b) .544/.2
c)1/2MVo^2 = MGH
H= Vo^2 / 2G

Answer 4

h = v0t - (1/2)at^2
0.2v0 - (1/2)(9.80665)(0.2)^2 = 0.544
v0 = 3.7 m/s
v1 = 3.7 - 0.2*9.80665 = 1.74 m/s
hf = (3.7^2)/2/9.80665 = 0.698 m or 69.8 cm