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RSB [31]
3 years ago
12

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.
Physics
1 answer:
Elena L [17]3 years ago
3 0
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.

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Generally distance covered along the horizontal is  

   D  =  v cos (40) *  2.07

=>   D  =  15 cos (40) *  2.07

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