Answer: The ball will hit the ground 6 seconds after it has ben thrown, at t = 6s.
Step-by-step explanation:
The equation for the height of the ball should be:
s(t) = (-16ft/s^2)*t^2 + 96ft/s*t
When s(t) = 0ft, means that the ball is in the ground
This represents the height of the ball as a function of t, the time in seconds.
We can see that at t = 0s:
s(0s) = (-16ft/s^2)*(0s)^2 + 96ft/s*(0s) = 0ft
Then at the time 0 seconds, the ball is in the ground, so we must look at the other root of the equation
(-16ft/s^2)*t^2 + 96ft/s*t = 0ft
To find it, we can use the Bhaskara equation, in this case is:
[tex]t = \frac{-96ft/s +- \sqrt{(96ft/s)^2 - 4*(-16ft/s^2)*0} }{-2*16ft/s^2} = \frac{-96ft/s^2 +-96ft/s^2}{-32ft/s^2}[/tex]
Then we have the two solutions:
t = (-96ft/s + 96ft/s)/(-32 ft/s^2) = 0s (the one that we already found)
And the other one is:
t = (-96ft/s - 96ft/s)/(-32ft/s^2) = 6s
Then:
s(6s) = 0ft
This means that the ball will hit the ground at t = 6 seconds.
