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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?

1 Answer

7 votes
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
User Ndou
by
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