Halliday, Fundamentals of Physics, 10e 1 system AUt PRINTER VERSION BACK Chapter

Question

Halliday, Fundamentals of Physics, 10e 1 system AUt PRINTER VERSION BACK Chapter 02, Problem 044 Your answer is partially correct. Try again. A startled armadillo leaps upward, rising 0.520 m in the first 0.180 s. (a) What is its initial speed as it leaves the ground? (b) what is its speed at the height of 0.520 m2 (c) How much higher does it go? Use g-9.81 m/s2 (a) Number T (b) Number jO.24 (c) Numbet-p.205 2.007 Units Tm/s SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE MATH HELD MATH HELD MATH HELP MATH HELD Question Attempts: 1 of 6 used AvA

Explanation / Answer

Here ,

time , t = 0.180 s

d = 0.520 m

a) let the initial speed is u

d = u * t - 0.50 gt^2

0.520 = u * 0.180 - 0.50 * 9.8 * 0.180^2

solving for u

u = 3.77 m/s

the initial speed is 3.77 m/s

b)

at this height

speed = u -gt

speed= 3.77 - 9.8 * 0.180

speed = 2 m/s

the speed at this height is 2 m/s

c)

more height reached = u^2/(2g)

more height reached = 2^2/(2 * 9.8)

more height reached = 0.204 m

it is go 0.204 m higher

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