Answer:
Using the equation given, we have to derive, because the derivative of a displacement functions results in velocity, and then we just calculate time, like this:
![s=-16t^{2}+v_{0}t+s_{0}](https://tex.z-dn.net/?f=s%3D-16t%5E%7B2%7D%2Bv_%7B0%7Dt%2Bs_%7B0%7D)
We know that
and ![v_{0}=128ft/sec](https://tex.z-dn.net/?f=v_%7B0%7D%3D128ft%2Fsec)
Replacing these values: ![s=-16t^{2} +128t+0](https://tex.z-dn.net/?f=s%3D-16t%5E%7B2%7D%20%2B128t%2B0)
Then, we derive: ![v=2(-16)t^{2-1} +128x^{1-1}\\ v=-32t+128t^{0} \\v=-32t+128](https://tex.z-dn.net/?f=v%3D2%28-16%29t%5E%7B2-1%7D%20%2B128x%5E%7B1-1%7D%5C%5C%20v%3D-32t%2B128t%5E%7B0%7D%20%5C%5Cv%3D-32t%2B128)
But, when the object reach the highest point, the velocity in that moment is zero. (v = 0)
So,
.
Therefore, the object takes 4 seconds long to reach the highest point. Also, this is a symmetrical movement (because is a quadratic function), this means that the object uses double of this time to reach the ground. So, <em>after 8 seconds, the object will be back at ground level. </em>
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To solve the second part of the problem, we have to solve a quadratic inequality. A height less than 128 feet can be expressed like: ![s](https://tex.z-dn.net/?f=s%3C128)
We know that ![s=-16t^{2} +128t](https://tex.z-dn.net/?f=s%3D-16t%5E%7B2%7D%20%2B128t)
Replacing this s expression on the inequality:
![-16t^{2}+128t](https://tex.z-dn.net/?f=-16t%5E%7B2%7D%2B128t%3C128)
Solving this inequality:
![-16t^{2} +128t-128](https://tex.z-dn.net/?f=-16t%5E%7B2%7D%20%2B128t-128%3C0%5C%5C)
Dividing by -16:
; which is easier too solve.
Therefore, solving this quadratic equation, the roots are = 6.82843 and
=1.17157. So, the answer that makes sense is =1.17157, because the total time to reach the highest point is 4 and 6 seconds is more than 4.
<em>Hence, at 1.2 seconds, the object will be in a position less than 128 feet.</em>
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