A) 436 ft
B) t = 5 sec
C) 5.60 s
Step-by-step explanation:
A)
The height of the ball at time t is given by the equation
![h(t)=-16t^2+500](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B500)
where
is the acceleration of the ball (acceleration of gravity, downward)
+500 is the initial height of the ball, at time t = 0
Here we want to find the height of the ball after 2 seconds, so at a time of
t = 2 s
Substituting into the equation, we find:
![h(2)=-16\cdot 2^2+500=436 ft](https://tex.z-dn.net/?f=h%282%29%3D-16%5Ccdot%202%5E2%2B500%3D436%20ft)
B)
Here we want to find the time it takes for the ball to fall to a height of 100 feet above the ground, so the time t at which
h(t) = 100 ft
As stated in the text of the question, the height of the ball at time t is given by
![h(t)=-16t^2+500](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B500)
Since
h(t) = 100, we have
![100=-16t^2+500](https://tex.z-dn.net/?f=100%3D-16t%5E2%2B500)
And solving for t we find:
![16t^2=400\\t^2=25\\t=\pm 5](https://tex.z-dn.net/?f=16t%5E2%3D400%5C%5Ct%5E2%3D25%5C%5Ct%3D%5Cpm%205)
So, the correct solution is the positive one:
t = 5 sec
C)
The ball reaches the ground when the height of the ball has became zero:
![h(t) = 0](https://tex.z-dn.net/?f=h%28t%29%20%3D%200)
The height of the ball at time t is given by
![h(t)=-16t^2+500](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B500)
And substituting
h(t) = 0
We get
![0=-16t^2+500](https://tex.z-dn.net/?f=0%3D-16t%5E2%2B500)
And solving the equation for t, we find the time t at which the ball reaches the ground:
![16t^2=500\\t^2=31.25\\t=\pm 5.6 s](https://tex.z-dn.net/?f=16t%5E2%3D500%5C%5Ct%5E2%3D31.25%5C%5Ct%3D%5Cpm%205.6%20s)
So, the correct solution is the positive one:
t = 5.60 s