Question

martin throws a ball straight up in the air. the equation h(t) = -16t^2 + 40t + 5 gives the height of the ball, in feet, t seconds after martin releases it.
how many seconds before the ball martin threw hits the ground?
show your work.
(pls answer asap!!)

Answer 1

Answer:

2.62 seconds

Step-by-step explanation:

Let

t ----> the time in seconds

h(t) ----> he height of the ball, in feet

we have

$h(t)=-16t^2+40t+5$

we know that

When the ball hits the ground, the height is equal to zero

so

$-16t^2+40t+5=0$

The formula to solve a quadratic equation of the form

$at^{2} +bt+c=0$

is equal to

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

in this problem we have

$-16t^2+40t+5=0$

so

$a=-16\\b=40\\c=5$

substitute in the formula

$t=\frac{-40\pm\sqrt{40^{2}-4(-16)(5)}} {2(-16)}$

$t=\frac{-40\pm\sqrt{1,920}} {-32}$

$t=\frac{-40+\sqrt{1,920}} {-32}=-0.12\ sec$

$t=\frac{-40-\sqrt{1,920}} {-32}=2.62\ sec$

therefore

The solution  is t=2.62 seconds