Question

An object is thrown upward at a speed of 96 feet per second by a machine from a height of 14 feet off the ground. The height of the object after t t seconds can be found using the equation h ( t ) = − 16 t 2 + 96 t + 14 h(t)=-16t2+96t+14. Give all numerical answers to 2 decimal places.

Answer 1

Answer:

$8t^2 -48 t -7=0$

$x =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

Where:

a= 8, b =-48, c = -7 and replacing we got:

$t_1 = \frac{48 - \sqrt{(-48)^2 -4(8)(-7)}}{2*8} = -0.1425$

$t_2 = \frac{48 + \sqrt{(-48)^2 -4(8)(-7)}}{2*8} = 6.142$

And since the time cannot be negative the correct answer for this cae would be t = 6.142 s

Step-by-step explanation:

If the question is: When will the object hit the ground at when the time is how many seconds?

We have the function for the heigth given by:

$h(t) = -16t^2 +96 t +14$

And we want to find the time where $h(t) =0$ so we can set like this the condition:

$0 =-16t^2 +96 t +14$

And we can rewrite the expression like this:

$16t^2 -96 t -14 =0$

We can divide both sides of the equation by 2 and we got:

$8t^2 -48 t -7=0$

And we can use the quadratic equation to solve for t like this:

$x =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

Where:

a= 8, b =-48, c = -7 and replacing we got:

$t_1 = \frac{48 - \sqrt{(-48)^2 -4(8)(-7)}}{2*8} = -0.1425$

$t_2 = \frac{48 + \sqrt{(-48)^2 -4(8)(-7)}}{2*8} = 6.142$

And since the time cannot be negative the correct answer for this cae would be t = 6.142 s

Answer 2

Step-by-step explanation:

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