Question

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?

Answer 1

Answer:

After 3 seconds

Step-by-step explanation:

Given

$h(t) = -5t^2 + 14t + 3$

Required

Time the drop will hit the ground

When the drop hits the ground,

$h(t) = 0$

So, we have:

$0 = -5t^2 + 14t + 3$

Rewrite as:

$5t^2 - 14t - 3 = 0$

Expand

$5t^2 + t - 15t - 3 = 0$

Factorize

$t(5t + 1) -3(5t + 1) = 0$

Factor out 5t + 1

$(t -3)(5t + 1) = 0$

Split

$t -3 = 0$ or $5t + 1 = 0$

Solve for t

$t = 3$ or $5t = -1$

Time cannot be negative.

So:

$t = 3$