A water balloon is tossed into the air with an upward velocity of 25 ft/s.

Question

Its height h(t) in ft after t seconds is given by the function h(t) = -16t^2 + 20t + 14.

1. After how many seconds will the balloon hit the ground?
2. What will the height be at t = 1 second?

Answer 1

After how many seconds will the balloon hit the ground

set h(t)=0 , factor and solve for "t"

(t+1/2)(t-7/4)=0

since time should be positive, solve the below eqt. to find your answer:

t-7/4= ? sec

1. What will the height be at t = 1 second?

replace "t" with 1 second in h(t)

h(1)= -16(1)^2+20(1)+14

h(1)= ? feet

Answer 2

h(t) = -16t^2 +20t + 14 =0

8t^2 -10t -7 = 0

(2t+1)(4t-7) =0

t = 7/4 = 1 3/4 seconds until it hits the ground

h(1) = -16+20+14 = 18 feet after seconds

However, your question may involve a typo. the equation assumes an initial velocity upwards of 20 feet per second, but you also state 25 feet per second. that would mean h(t)=-16t^2 +25t +14, with a different solution.