Answer:
Maximum height is 29 ft
Time to reach the ground is 2.60 seconds
Step-by-step explanation:
We are given equation of height as
![h(t)=-16t^2+40t+4](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B40t%2B4)
Maximum height:
we know that at maximum height
velocity =0
so, we will find derivative
and then we can set it to 0
and we solve for t
![h'(t)=-32t+40](https://tex.z-dn.net/?f=h%27%28t%29%3D-32t%2B40)
now, we can set it to 0
and then we can solve for t
![h'(t)=-32t+40=0](https://tex.z-dn.net/?f=h%27%28t%29%3D-32t%2B40%3D0)
![-32t=-40](https://tex.z-dn.net/?f=-32t%3D-40)
![t=\frac{5}{4}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B5%7D%7B4%7D)
![t=1.25](https://tex.z-dn.net/?f=t%3D1.25)
now, we can plug t into height equation
![h(\frac{5}{4})=-16(\frac{5}{4})^2+40(\frac{5}{4})+4](https://tex.z-dn.net/?f=h%28%5Cfrac%7B5%7D%7B4%7D%29%3D-16%28%5Cfrac%7B5%7D%7B4%7D%29%5E2%2B40%28%5Cfrac%7B5%7D%7B4%7D%29%2B4)
![h(1.25)=29](https://tex.z-dn.net/?f=h%281.25%29%3D29)
Time to reach the ground:
we know that at ground
height=0
so, we can set h(t)=0
and then we can solve for t
![h(t)=-16t^2+40t+4=0](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B40t%2B4%3D0)
we can use quadratic formula
![t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
![t=\frac{-40\pm \sqrt{40^2-4\left(-16\right)4}}{2\left(-16\right)}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-40%5Cpm%20%5Csqrt%7B40%5E2-4%5Cleft%28-16%5Cright%294%7D%7D%7B2%5Cleft%28-16%5Cright%29%7D)
![t=-\frac{\sqrt{29}-5}{4},\:t=\frac{5+\sqrt{29}}{4}](https://tex.z-dn.net/?f=t%3D-%5Cfrac%7B%5Csqrt%7B29%7D-5%7D%7B4%7D%2C%5C%3At%3D%5Cfrac%7B5%2B%5Csqrt%7B29%7D%7D%7B4%7D)
![t=-0.09629,t=2.596](https://tex.z-dn.net/?f=t%3D-0.09629%2Ct%3D2.596)
Since, time can never be negative
so, we will only consider positive time
![t=2.596](https://tex.z-dn.net/?f=t%3D2.596)