contestada

The vertical position y (in feet) of a rock t seconds after it was dropped from a cliff is given by the formula y equals short dash 16 t squared plus 4 t plus 380. The base of the cliff corresponds to y equals 0. After how many seconds will the rock hit the ground at the base of the cliff?

Respuesta :

Answer: t = 4.87s

Step-by-step explanation: The vertical position y is given by

[tex]y = -16t^{2}+380[/tex]

If y at the base of the cliff equals 0, then equation is

[tex]-16t^{2}+380=0[/tex]

[tex]-16t^{2}=-380[/tex]

[tex]t^{2}=\frac{-380}{-16}[/tex]

[tex]t=\sqrt{\frac{380}{16}}[/tex]

[tex]t=\sqrt{23.75}[/tex]

t = 4.87

Since it is time we want, the negative value of the square root can be ignored. So, at the base of the cliff, the rock will hit the ground after 4.87 seconds.

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