Question

Suppose you have a water balloon launcher. The balloon is 3ft high when it leaves the launcher. Use the equation 0=-16t^2+38.8t+3 to find the number of seconds t that the balloon is in the air.

Answer 1

`16t^2 -38.8 t -3 =0`

And we can use the quadratic formula given by:

`t = (-b \pm √(b^2 -4ac))/(2a)`

Where:

`a = 16, b=-38.8, c = -3`

And replacing we got:

`t = (38.8 \pm √((-38.8)^2 -4(16)(-3)))/(2*16)`

And after solve we got two solutions:

`t_1 = 2.5 s`

And
`t_2 =-0.075 s`

Since the time can't be negative the correct option for this case would be
`t =2.5 s`

Explanation:

For this case we have the following function for the height:

`h(t) = -16t^2 +38.8t +3`

And we want to find how many seconds t that the balloon is in the air since is released from 3ft above, so we want to find the time t in order to h(t) =0

`0= -16t^2 +38.8t +3`

We can rewrite the last expression like this:

`16t^2 -38.8 t -3 =0`

And we can use the quadratic formula given by:

`t = (-b \pm √(b^2 -4ac))/(2a)`

Where:

`a = 16, b=-38.8, c = -3`

And replacing we got:

`t = (38.8 \pm √((-38.8)^2 -4(16)(-3)))/(2*16)`

And after solve we got two solutions:

`t_1 = 2.5 s`

And
`t_2 =-0.075 s`

Since the time can't be negative the correct option for this case would be
`t =2.5 s`