Question

a water balloon is tossed into the air with an upward velocity of 25 ft/s its height h(t) in ft after t seconds is given by the function h(t) = -16t^2+25t+3 after how many seconds will the balloons hit the ground?

Answer 1

when the height is 0 that is when it hits the ground
solve
0=-16t²+25x+3
cant factor so use quadratic formula

for
0=ax²+bx+c

`x=(-b+/-√(b^2-4ac))/(2a)`

so for
0=-16t²+25t+3
a=-16
b=25
c=3
so


`x=(-25+/-√(25^2-4(-16)(3)))/(2(-16))`

`x=\frac{-25+/-\frac{625+192}}{-32}`

`x=\frac{-25+/-\frac{817}}{-32}`
it has to be a positive time

so it will hit the ground after
`(25+√(817))/(32)` seconds