Question

The equation h(t) = -16t² + 80t + 64 represented the height, in feet, of a potato t seconds after it has been launched.

Answer 1

Answer:

Part A) The potato hit the ground at t=5.70 seconds (see the explanation)

Part B) The potato is 40 feet off the ground at the time t=5.28 seconds (see the explanation)

Step-by-step explanation:

we have

$h(t)=-16t^2+80t+64$

where

h(t) is the height of a potato in feet

t is the time in seconds

Part A)  Write an equation that can be solved to find when the potato hits the ground. Then solve the equation

we know that

When the potato hit the ground, the value of h(t) must be equal to zero

so

For h(t)=0

$-16t^2+80t+64=0$

Solve the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-16t^2+80t+64=0$

so

$a=-16\\b=80\\c=64$

substitute in the formula

$t=\frac{-80\pm\sqrt{80^{2}-4(-16)(64)}} {2(-16)}$

$t=\frac{-80\pm\sqrt{10,496}} {-32}$

$t=\frac{-80+\sqrt{10,496}} {-32}=-0.70$

$t=\frac{-80-\sqrt{10,496}} {-32}=5.70$

therefore

The potato hit the ground at t=5.70 seconds

Part B) Write an equation that can be solved to find when the potato is 40 feet off the ground. Then solve the equation

For h(t)=40 ft

substitute in the quadratic equation

$-16t^2+80t+64=40$

$-16t^2+80t+24=0$

Solve the quadratic equation

we have

$a=-16\\b=80\\c=24$

substitute in the formula

$t=\frac{-80\pm\sqrt{80^{2}-4(-16)(24)}} {2(-16)}$

$t=\frac{-80\pm\sqrt{7,936}} {-32}$

$t=\frac{-80+\sqrt{7,936}} {-32}=-0.28$

$t=\frac{-80-\sqrt{7,936}} {-32}=5.28$

therefore

The potato is 40 feet off the ground at the time t=5.28 seconds