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A ball is thrown downward from the top of a 120-foot building with an initial velocity of 20 feet per second. The
height of the ball h in feet after t seconds is given by the equation h= - 16t²-20t+120. How long after the
ball is thrown will it strike the ground?

Respuesta :

Medunno13

Answer:

[tex]t=\frac{-5+\sqrt{505}}{8}[/tex]

Step-by-step explanation:

The ball strikes the ground when h = 0.

[tex]-16t^2 - 20t + 120 = 0 \\ \\ 4t^2 + 5t - 30=0 \\ \\ t=\frac{-5 \pm \sqrt{5^{2}-4(4)(-30)}}{4(2)} \\ \\ t=\frac{-5 \pm \sqrt{505}}{8}[/tex]

However, as time most be positive, we only consider the positive case.

So,

[tex]t=\frac{-5+\sqrt{505}}{8}[/tex]

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