Correct option:
d. 1.64
An object is launched vertically upward from a 25 foot platform and is modeled by the following function:
[tex]s(t)=-16t^2+11t+25[/tex]
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 [tex]s(t)=0[/tex] so:
[tex]-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[/tex]
Since time can't be negative, we only choose the positive value. therefore:
The object fall back to the ground after 1.64 seconds
Projectile motion: https://brainly.com/question/12554459
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