Question

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The height of an object launched upward from ground level is given by the function s(t) = -4.9t2 + 78.4t, where s(t) is the height in meters and t is the time in seconds. Arrange the time intervals in ascending order of the object's magnitude of average velocity over the interval.

Answer 1

Answer:

the distance of  the object before it  hits the ground after launch is t = 1.592 seconds

Step-by-step explanation:

The height of an object launched upward from ground level is given by the function s(t) = -4.9t² +7.8t

when the object reach the ground then s(t) = 0

So;  -4.9t² +7.8t = 0

-4.9t² +7.8t + 0 = 0

By using the quadratic equation

$\mathbf{\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}}$

where; a = -4.9 ; b = 7.8 and c = 0

$\mathbf{=\dfrac{-(7.8) \pm \sqrt{(7.8)^2-4(-4.9)(0)} }{2(-4.9)}}$

$\mathbf{=\dfrac{-(7.8) \pm \sqrt{(60.84} }{-9.8}}$

$\mathbf{=\dfrac{-(7.8) + \sqrt{(60.84} }{-9.8} \ \ \ \ OR \ \ \ \ \mathbf{\dfrac{-(7.8) - \sqrt{(60.84} }{-9.8}}}$

$\mathbf{=0 \ \ \ OR \ \ \ 1.592 }}$

Hence;  the distance of  the object before it  hits the ground after launch is t = 1.592 seconds