SOLUTION: A water fountain jet shoots short spurts of water over a walkway. The water spurts reach a maximum height, then come down into a pond of water on the other side of the walkway.

Question 1161371: A water fountain jet shoots short spurts of water over a walkway. The water spurts reach a maximum height, then come down into a pond of water on the other side of the walkway. The height above the jet, h, of a spurt of water t seconds after leaving the jet can be found by the function h(t)= -16t^2d+20t. Find the time it takes for the spurt of water to return to the jet's height; that is, when h(t)=0.
Thank you
Answer by ikleyn(52754) (Show Source): You can put this solution on YOUR website!
.


All you need is to solve this quadratic equation


    -16t^2 + 20t = 0,   or, equivalently

     16t^2 - 20t = 0.


Factorize


    4t*(4t-5) = 0.


One root is t= 0; it is the starting moment; and obviously, you need another root  t=  seconds = 1.25 seconds.    ANSWER

Solved.

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SOLUTION: A water fountain jet shoots short spurts of water over a walkway. The water spurts reach a maximum​ height, then come down into a pond of water on the other side of the walkway.

SOLUTION: A water fountain jet shoots short spurts of water over a walkway. The water spurts reach a maximum height, then come down into a pond of water on the other side of the walkway.