Answer :
Answer:
t = 2.64 or [tex]t = \frac{3\ + \sqrt{329} }{8 }[/tex]
Step-by-step explanation:
v(t) = [tex]\int\limits {-32} \, dx[/tex]
v(t) = -32t + C
when t = 0, the velocity is 12
12 = -32(0) + C
C = 12
v(t) = -32t + 12
now finding position
[tex]\int\limits {(-32t+12)} \, dx[/tex]
[tex]x(t) = -16t^2 +12t + C[/tex]
when t = 0, position is 80
80 = 0 + C, C = 80
[tex]x(t) = -16t^2 +12t + 80[/tex]
now finding when position = 0
0 = [tex]-16t^2 +12t + 80[/tex]
0 = [tex]4t^2-3t-20[/tex]
[tex]t = \frac{3\frac{+}{} \sqrt{9 + 320} }{8 }[/tex]
t = -1.89, 2.64
t>0, so t = 2.64
The required value is [tex]\frac{3+\sqrt{329}}{8}[/tex].
Integration:
The integration denotes the summation of discrete data. The integral is calculated to find the functions which will describe the area, displacement, volume, that occurs due to a collection of small data, which cannot be measured singularly.
Let [tex]v(t)[/tex] denote the velocity of the package at the time [tex]t[/tex], and let [tex]s[/tex] denotes the height at the time [tex]t[/tex].
Acceleration due to gravity is given [tex]32ft/sec^{2}[/tex]. That is we have,
[tex]\frac{dv}{dt} =-32[/tex] [Negative because 'g' acts in the direction of decreasing s]
Initial condition: [tex]v(0)=12[/tex]
[tex]dv=-32dt\\v=-32t+C\\12=-32(0)+C\\C=12[/tex]
So, [tex]v=-32t+12[/tex]
Since velocity is the derivative of height,
[tex]\frac{ds}{dt} =-32t+12[/tex]
Initial condition: [tex]S(0)=80[/tex]
[tex]ds=(-32t+12)dt\\S=-16t^{2} +12t+C\\80=-16(0)+12(0)+C\\C=80[/tex]
The package height above the ground at time "[tex]t[/tex]" is,
[tex]S=-16t^{2} +12t+80[/tex]
When package touches the ground,
[tex]S=-16t^{2} +12t+80=0\\4t^{2} -3t-20=0\\t=\frac{-(-3)\pm \sqrt{9+320} }{2(4)} \\=\frac{3\pm \sqrt{329} }{8} \\=\frac{3+ \sqrt{329} }{8}[/tex]
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