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An engineer is designing a fountain that shoots out drops of water. The nozzle from which the water is launched is 3 meters above the ground. It shoots out a drop of water at a vertical velocity of 14 meters per second. Function h models the height in meters, h, of a drop of water t seconds after it is shot out from the nozzle. The function is defined by the equation h(t)=−5t2+14t+3. How many seconds until the drop of water hits the ground?

Respuesta :

MrRoyal

Answer:

After 3 seconds

Step-by-step explanation:

Given

[tex]h(t) = -5t^2 + 14t + 3[/tex]

Required

Time the drop will hit the ground

When the drop hits the ground,

[tex]h(t) = 0[/tex]

So, we have:

[tex]0 = -5t^2 + 14t + 3[/tex]

Rewrite as:

[tex]5t^2 - 14t - 3 = 0[/tex]

Expand

[tex]5t^2 + t - 15t - 3 = 0[/tex]

Factorize

[tex]t(5t + 1) -3(5t + 1) = 0[/tex]

Factor out 5t + 1

[tex](t -3)(5t + 1) = 0[/tex]

Split

[tex]t -3 = 0[/tex] or [tex]5t + 1 = 0[/tex]

Solve for t

[tex]t = 3[/tex] or [tex]5t = -1[/tex]

Time cannot be negative.

So:

[tex]t = 3[/tex]

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