Question

A ball is launched upward from a height 40 feet above ground level. The ball’s height at t seconds is given by -16t^2=128=40 .

Answer 1

The ball will be at 100 feet at time x=0.5 sec and at time x=7.5 sec

Explanation:

The correct question is

A ball is launched upward from a height 40 feet above ground level. the ball's height at t seconds is given by h(t)=-16t^2+128t+40 At what time(s) will the ball be at 100 feet?

we have

`h(t)=-16t^(2)+128t+40`

so

For h(t)=100 ft

substitute in the equation and solve for x

`-16t^(2)+128t+40=100`

`-16t^(2)+128t-60=0`

The formula to solve a quadratic equation of the form

`ax^(2) +bx+c=0`

is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`-16t^(2)+128t-60=0`

so

`a=-16\\b=128\\c=-60`

substitute in the formula

`x=\frac{-128(+/-)\sqrt{128^(2)-4(-16)(-60)}} {2(-16)}`

`x=\frac{-128(+/-)√(12,544)} {-32}`

`x=\frac{-128(+/-)112} {-32}`

`x=\frac{-128(+)112} {-32}=0.5`

`x=\frac{-128(-)112} {-32}=7.5`

therefore

The ball will be at 100 feet at time x=0.5 sec and at time x=7.5 sec