Help please! Matthew launches a ball is at 34 meters per sec from a 28 meter tall platform. The function of the ball's height (h) at time (*) seconds after launch is
h(t) = -12t^2 + 34t + 28.

17 seconds

Step-by-step explanation:

h(t) represents height, we want to know when the ball will return to a height of 28 feet. So our equation is..

28 = -2t² + 34t + 28

Now solve for t...

0 = -2t² + 34t (subtract 28 from both sides)

0 = t² - 17t (divide both sides by -2)

0 = t(t - 17) (factor out a t)

So

t = 0

or

t - 17 = 0

t = 17

0 seconds is the initial height. At 17 seconds, the ball will return to that height

i hope this work for you

attachment?

step-by-step explanation: