Question

A water balloon is catapulted into the air so that its height H , in meters, after T seconds is h=-4.9t=27t=2.4.Help me solve these questions?

A.) How high is the ball after 1 second?
B.)What is the maximum height of the balloon?
C.)When will the balloon burst as it hits the ground?

Answer 1

A) `h(1)=24.5m`
B) `h(2.755)=39.59m`
C) `x=5.60"seconds"`

Explanation:

I'll assume that `h=-4.9t=27t=2.4` should be `h=-4.9t^2+27t+2.4`

A)
Solve in terms of `t=(1)`
`h(1)=-4.9(1)^2+27(1)+2.4` `color(blue)("Add")`
`h(1)=color(red)(24.5m)`

B)
Vertex formula is `((-b)/(2a), h((-b)/(2a)))`
Remember: `ax^2+bx+c`
Vertex: `(-27)/(2(-4.9)) = 2.755` `color(blue)("Solve")`
`h((-b)/(2a))=h(2.755)` `color(blue)("Plug 2.755 into t in the original equation")`
`h(2.755)=-4.9(2.755)^2+27(2.755)+2.4` `color(blue)("Solve")`
`h(2.755)=color(red)(39.59m)`

C)
Find the `"x-intercepts"` using the quadratic formula:
`(-b±sqrt((b)^2-4ac))/(2a)`
`(-(27)±sqrt((27)^2-4(-4.9)(2.4)))/(2(-4.9)` `color(blue)("Solve")`
`(-27±27.86)/-9.8` `color(blue)("Determine which x-intercept is logical in this situation")`

`(-27+27.86)/-9.8=-.0877" seconds"`

`(-27-27.86)/-9.8=5.5979" seconds"`

A negative value in terms of seconds would not make sense in this problem, therefore the answer is `color(red)(x=5.60" seconds"))`