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

Answer:

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

Step-by-step 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

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

$x=\frac{-128(+/-)\sqrt{12,544}} {-32}$

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

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

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

therefore

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