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An object is dropped from 24 feet below the tip of the pinnacle atop a 1468-ft tall building. The height h of the object after t seconds is given by the equatior h= - 16t2 + 1444. Find how many seconds pass before the object reaches the ground. seconds pass before the object reaches the ground. (Type an integer or a decimal.)​

Respuesta :

Answer:

9.5 seconds pass before the object reaches the ground.

Step-by-step explanation:

Height of the ball:

The height of the ball after t seconds is given by the following equation:

[tex]h(t) = -16t^2 + 1444[/tex]

Find how many seconds pass before the object reaches the ground.

This is t for which h(t) = 0. So

[tex]h(t) = -16t^2 + 1444[/tex]

[tex]-16t^2 + 1444 = 0[/tex]

[tex]16t^2 = 1444[/tex]

[tex]t^2 = \frac{1444}{16}[/tex]

[tex]t^2 = 90.25[/tex]

[tex]t = \pm \sqrt{90.25}[/tex]

Since it is time, we only take the positive value.

[tex]t = 9.5[/tex]

9.5 seconds pass before the object reaches the ground.

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