Question

Derek kicks a soccer ball off the ground and in the air, with an initial velocity of 31 feet per second. Using the formula H(t) = −16t2 + vt + s, what is the maximum height the soccer ball reaches? (5 points) 14.2 feet 14.6 feet 15.0 feet 15.3 feet

Answer 1

General Idea:

If we have a quadratic function of the form $f(x)=ax^{2} +bx+c$, then the function will attain its maximum value only if a < 0 & its maximum value will be at $x=-\frac{b}{2a}$.

Applying the concept:

Comparing the function$f(x)=ax^{2} +bx+c$ with the given function$H(t)=-16t^{2} +31t+s$, we get$a = -16$, $b = 31$ and $c=0$ because initial height of the ball s will be 0.

The maximum height of the soccer ball will occur at $t=\frac{-b}{2a}=\frac{-31}{2(-16)} = \frac{-31}{-32}=0.96875 seconds$

The maximum height is found by substituting $t=0.96875$ in the function as below:

$H(0.96875)=-16(0.96875)^2+31(0.96875)$

Conclusion:

The maximum height the soccer ball reaches is 15.0 feet approximately!