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A water balloon is tossed into the air with an upward velocity of 25 ft/s. Its height h(t) in ft after t seconds is given by the function h(t) = − 16t2 + 25t + 3. Show your work.

After how many seconds will the balloon hit the ground? (hint: Use the quadratic formula)


What will the height be at t = 1 second
PLEASE HELP,,STUCK :(

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1) [tex]h (t) = -16t^2 + 25t + 3 \\ \\h(t) = 0 \\ \\ 0 = -16t^2 + 25t + 3 \\ \\ quadratic \ formula \\ \\ a = -16 \ , \ b = 25 \ , c = 3 \\ \\ t = \frac{-25 \pm \sqrt{25^2-4(-16)(3)} }{2(-16)} \\ \\ \frac{-25 \pm \sqrt{625+192} }{-32} \\ \\ \frac{-25 \pm \sqrt{817} }{-32} \\ \\ \frac{-25 \pm 28.5832}{-32} \\ \\ t = \frac{-25 - 28.5832}{-32} \\ \\ t = \frac{-25 + 28.5832}{-32} \\ \\ \frac{53.5832}{32} \\ \\ \frac{53.5832}{32} \\ \\ 1.66447[/tex]

The water ballon will hit the ground after 1.66447 seconds.

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2) [tex]t = 1 \\ \\ h(1) = -16(1)^2 + 25(1) + 3 \\ \\ -16 + 25 + 3 \\ \\ 12 [/tex]

The height will be 12 feet if t = 1 second.

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