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) = − 16t2 + 25t + 3. Show your work.

After how many seconds will the balloon hit the ground? (hint: Use the quadratic formula)

What will the height be at t = 1 second

The balloon hits the ground means -16t2+25t+3=0
After you solve the equation, there are 2 solution ~1.67 or ~-0.11 (the time can not be negative) so the solution is 1.67s
20.substitute t=1 to the function -16(1)2+25(1)+3=-16+25+3=12 ft. So the height is 12ft when t=1

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aachen aachen · Answer 2 · 2019-09-09T09:10:05-04:00

Step-by-step explanation:

Given that,

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 as :

\(h(t)=-16t^2+25t+3\)...................(1)

(a) we need to find the time when the balloon hits the ground. When it hits the ground h(t) = 0

\(-16t^2+25t+3=0\)

Using above equation we get the value of t, t = 1.674 seconds

So, after 1.674 seconds the balloon hits the ground.

(b) We need to find the height at t = 1 seconds. Put the value of t = 1 seconds in equation (1) as :

\(h(1)=-16(1)^2+25(1)+3\)

h = 12 feet

So, at t = 1 seconds the height of the balloon is 12 feets.