2 years ago

Which point is a solution to the equation?

Mathematics

1 answer:

ryzh [129]2 years ago

5 0

I think point b I’m not sure

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An object is launched vertically upward from a 25 foot platform and is modeled by the function s(t)=−16t2+11t+25.

ch4aika [34]

Correct option:

d. 1.64

<h2>Explanation:</h2>

An object is launched vertically upward from a 25 foot platform and is modeled by the following function:

$s(t)=-16t^2+11t+25$

So this is the equation of a parabola because the object is a projectile so it follows the Projectile Motion.

The object fall back to the ground when $s(t)=0$ so:

$-16t^2+11t+25=0 \\ \\ \\ Using \ quadratic \ formula: \\ \\ t_{12}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ a=-16 \\ \\ b=11 \\ \\ c=25 \\ \\ \\ t_{12}=\frac{-11\pm \sqrt{11^2-4(-16)(25)}}{2(-16)} \\ \\ t_{12}=\frac{-11\pm 41.48}{-32} \\ \\ t_{1}=\frac{-11+ 41.48}{-32}=-0.95s \\ \\ t_{2}=\frac{-11- 41.48}{-32}=1.64s$