Answer:
The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>
Step-by-step explanation:
Given:
Initial speed of debris is, ![s=120\ ft/s](https://tex.z-dn.net/?f=s%3D120%5C%20ft%2Fs)
The height 'h' of the debris above the ground is given as:
![h(t)=-16t^2+120t](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B120t)
As per question,
. Therefore,
![56=-16t^2+120t](https://tex.z-dn.net/?f=56%3D-16t%5E2%2B120t)
Rewriting the above equation into a standard quadratic equation and solving for 't', we get:
![-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0](https://tex.z-dn.net/?f=-16t%5E2%2B120t-56%3D0%5C%5C%5Ctextrm%7BDividing%20by%20-8%20throughout%2C%20we%20get%7D%5C%5C%5Cfrac%7B-16%7D%7B-8%7Dt%5E2%2B%5Cfrac%7B120%7D%7B-8%7Dt-%5Cfrac%7B56%7D%7B-8%7D%3D0%5C%5C2t%5E2-15t%2B7%3D0)
Using quadratic formula to solve for 't', we get:
![t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%5C%5C%5C%5Ct%3D%5Cfrac%7B-%28-15%29%5Cpm%20%5Csqrt%7B%28-15%29%5E2-4%282%29%287%29%7D%7D%7B2%282%29%7D%5C%5C%5C%5Ct%3D%5Cfrac%7B15%5Cpm%20%5Csqrt%7B225-56%7D%7D%7B4%7D%5C%5C%5C%5Ct%3D%5Cfrac%7B15%5Cpm%5Csqrt%7B169%7D%7D%7B4%7D%5C%5C%5C%5Ct%3D%5Cfrac%7B15%5Cpm%2013%7D%7B4%7D%5C%5C%5C%5Ct%3D%5Cfrac%7B15-13%7D%7B4%7D%5C%20or%5C%20t%3D%5Cfrac%7B15%2B13%7D%7B4%7D%5C%5C%5C%5Ct%3D%5Cfrac%7B2%7D%7B4%7D%5C%20or%5C%20t%3D%5Cfrac%7B28%7D%7B4%7D%5C%5C%5C%5Ct%3D0.5%5C%20s%5C%20or%5C%20t%3D7%5C%20s)
Therefore, the debris will reach a height of 56 ft twice.
When time
during the upward journey, the debris is at height of 56 ft.
Again after reaching maximum height, the debris falls back and at
, the height is 56 ft.