Given that Martin throws a ball straight up in the air.
The equation
gives the height of the ball, in feet, t seconds after martin releases it.
We need to determine the time that it takes the ball to hit the ground.
<u>Time taken:</u>
To determine the time 't', let us equate h(t) = 0 in the equation ![h(t)=-16t^2+40t+5](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B40t%2B5)
Thus, we have;
![0=-16t^2+40t+5](https://tex.z-dn.net/?f=0%3D-16t%5E2%2B40t%2B5)
Switch sides, we get;
![-16 t^{2}+40 t+5=0](https://tex.z-dn.net/?f=-16%20t%5E%7B2%7D%2B40%20t%2B5%3D0)
Now, we shall solve the equation using the quadratic formula.
Thus, we have;
![t=\frac{-40 \pm \sqrt{40^{2}-4(-16) 5}}{2(-16)}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-40%20%5Cpm%20%5Csqrt%7B40%5E%7B2%7D-4%28-16%29%205%7D%7D%7B2%28-16%29%7D)
Solving, we get,
![t=\frac{-40 \pm \sqrt{1600+320}}{-32}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-40%20%5Cpm%20%5Csqrt%7B1600%2B320%7D%7D%7B-32%7D)
![t=\frac{-40 \pm \sqrt{1920}}{-32}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-40%20%5Cpm%20%5Csqrt%7B1920%7D%7D%7B-32%7D)
![t=\frac{-40 \pm 8\sqrt{30}}{-32}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-40%20%5Cpm%208%5Csqrt%7B30%7D%7D%7B-32%7D)
![t=\frac{8(-5 \pm \sqrt{30})}{-32}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B8%28-5%20%5Cpm%20%5Csqrt%7B30%7D%29%7D%7B-32%7D)
Cancelling the common terms, we get,
![t=\frac{-5 \pm \sqrt{30}}{-4}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-5%20%5Cpm%20%5Csqrt%7B30%7D%7D%7B-4%7D)
Thus, the roots of the equation
and ![t=\frac{-5 - \sqrt{30}}{-4}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-5%20-%20%5Csqrt%7B30%7D%7D%7B-4%7D)
Simplifying the roots, we get,
and ![t=\frac{-5 -5.477}{-4}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-5%20-5.477%7D%7B-4%7D)
and ![t=2.619](https://tex.z-dn.net/?f=t%3D2.619)
Since, t cannot take negative values, then ![t=2.619](https://tex.z-dn.net/?f=t%3D2.619)
Hence, It takes 2.6 seconds for the ball to hit the ground.