Question
A startled armadillo leaps upward, rising 0.644 m in 0.170 s.
a. What is its initial speed?
b. What is its speed at this height?
c. How much higher does it go?
For a. 0 is not the answer and neither is 3.79 m/s (product of .644m*.170s)
a. What is its initial speed?
b. What is its speed at this height?
c. How much higher does it go?
For a. 0 is not the answer and neither is 3.79 m/s (product of .644m*.170s)
Answers
Henry
a. h = Vo*t + 0.5g*t^2 = 0.644 m.
Vo*0.17 - 4.9*0.17^2 = 0.644.
0.17Vo = 0.644 + 0.142 = 0.786.
Vo = 4.62 m/s.
b. V=Vo + g*t=4.62 - 9.8*0.17 = 2.95 m/s.
c. Vo^2 = Vo^2 + 2g*h = 0.
h = -Vo^2/2g = -(4.62^2)/-19.6 = 1.09 m.
1.09-0.644 = 0.446 m higher.
Vo*0.17 - 4.9*0.17^2 = 0.644.
0.17Vo = 0.644 + 0.142 = 0.786.
Vo = 4.62 m/s.
b. V=Vo + g*t=4.62 - 9.8*0.17 = 2.95 m/s.
c. Vo^2 = Vo^2 + 2g*h = 0.
h = -Vo^2/2g = -(4.62^2)/-19.6 = 1.09 m.
1.09-0.644 = 0.446 m higher.