Mathematics: 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!!)

-7x - 10y = 40-7x + 7x - 10y = 40 + 7x -10y = 40 + 7x -10 -10 y = -4 - ⁷/₁₀x...

Martin throws a ball straight up in the air. the

3kp

18.01.2022

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

-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

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

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

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