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An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.9t? + 20t + 65. What is the object's maximum height? The numeric answer only, rounded to the nearest meter.

Respuesta :

tramserran

Answer: 85 meters

Step-by-step explanation:

The maximum height is the y-value of the vertex.

h(t) = -4.9t² + 20t + 65

      a=-4.9   b=20  c=65

[tex]t=\dfrac{-b}{2a} = \dfrac{-(20)}{2(-4.9)} =\dfrac{-20}{-9.8}=2[/tex]

h(2) = -4.9(2)² + 20(2) + 65

      = -19.6 + 40 + 65

      = 85.4

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