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

After 3 seconds

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`