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Studentka2010 [4]
2 years ago
13

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 secon

ds after martin releases it.
how many seconds before the ball martin threw hits the ground?
show your work.
(pls answer asap!!)
Mathematics
1 answer:
Lisa [10]2 years ago
4 0

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

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