When the ball reaches the ground, its height is 0.
We set the height function equal to zero and solve for t, the time.
h = -16t^2 + 250
-16t^2 + 250 = 0
16t^2 - 250 = 0
[tex] (4t + \sqrt{250})(4t - \sqrt{250}) = 0 [/tex]
[tex] 4t + \sqrt{250} = 0 [/tex] or [tex] 4t - \sqrt{250} = 0 [/tex]
[tex] 4t = -\sqrt{250} [/tex] or [tex] 4t = \sqrt{250} [/tex]
[tex] t = -\dfrac{\sqrt{250}}{4} [/tex] or [tex] t = \dfrac{\sqrt{250}}{4} [/tex]
We discard the negative solution since time must be positive.
The positive solution is
[tex] t = \dfrac{\sqrt{250}}{4} \approx 4.0 [/tex]
Answer: 4.0 seconds
