Question

2. An object is thrown upward from the top of a 240 – foot cliff with a velocity of 12 feet per second. The height h in feet of the object after t second is h = -16t² + 12t + 240. How long after the object is thrown will it strike the ground? Round to the nearest tenth of a second.

Answer 1

Answer:

286t² answer, in tenth second 4000phs

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Answer 2

Given:

The height h in feet of the object after t second is h = -16t² + 12t + 240.

To find:

The object hits the ground approximately seconds.

Step-by-step explanation:

When object hits the ground, the height is 0. substitute 0 for h.

$h=-16t^{2} +12t+240$

$0=-16t^{2} +12t+240$

Quadratic formula,

$x=-b$±$\frac{\sqrt{b^{2}-4ac } }{2a}$ , in this case, the variable is t rather than x.

$a=-16,b=12,c=240$

$t=-(12)$±$\frac{\sqrt{12^{2}-4(-16)(240) } }{2(-16)}$

$t=-12$±$\frac{\sqrt{15504} }{-32}$

t is approximately$-15.54$ or $8.46.$

The object hits the ground approximately$8.46$ seconds after being thrown.