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3 years ago

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A startled armadillo leaps upward, rising 0.532 m in the first 0.202 s. (a) What is its initial speed as it leaves the ground? (

b) What is its speed at the height of 0.532 m? (c) How much higher does it go? Use g=9.81 m/s2.

1 answer:

MrMuchimi3 years ago

7 0

A) $d=v_0t+\frac{1}{2}at^2$
0.532 m = 0.202v_0 + ½(-9.81 m/s²)(0.202 s)²
v_0 = 3.62 m/s

B) $d = \frac{1}{2}(v+v_0)t$
0.532 m = ½(v + 3.62 m/s)(0.202) s
v = 1.65 m/s

C) $v^2=v_0^2+2ad$
At the highest point v = 0, so
0 = (1.65 m/s)^2 + 2(-9.81 m/s²)d
d = 0.139 m

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