Question 1197205
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t = number of seconds
h(t) = height of the water balloon


Let's determine what time value(s) apply when the height is 10 feet.


Replace h(t) with 10 and get everything to one side
h(t) = -16t^2+35t+5
10 = -16t^2+35t+5
0 = -16t^2+35t+5-10
0 = -16t^2+35t-5
-16t^2+35t-5 = 0


Consider the general form at^2+bt+c = 0
We have these coefficients
a = -16
b = 35
c = -5
in which we plug into the quadratic formula
{{{t = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{t = (-35+-sqrt((35)^2-4(-16)(-5)))/(2(-16))}}}


{{{t = (-35+-sqrt(905))/(-32)}}}


{{{t = (-35+-  30.083218)/(-32)}}}


{{{t = (-35+30.083218)/(-32)}}} or {{{t = (-35-30.083218)/(-32)}}}


{{{t = (-4.916782)/(-32)}}} or  {{{t = (-65.083216)/(-32)}}}


{{{t = 0.153649}}} or  {{{t = 2.033851}}}
The decimal values are approximate.


The water balloon reaches a height of 10 feet at approximately 0.153649 seconds. This is when the balloon is going upward.


It comes back down to get back to 10 feet at around 2.033851 seconds


The balloon is more than 10 feet off the ground for the interval <font color=red>0.153649 < t < 2.033851</font>
Basically anything between 0.153649 seconds and 2.033851 seconds, excluding each endpoint.


You can use graphing technology like Desmos or GeoGebra to visually confirm the answers. 
Be sure to use x in place of t, and use y in place of h(t)


Round each approximate decimal value according to the instructions your teacher provides.
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