Question

numeric 1 point a soccer ball is kicked from the ground with an initial upward velocity of 90 feet per second. the equation h(t) = -16t² + 90t gives the height h of the ball at t seconds. what is the maximum height of the ball? (round to the nearest hundredth.) answer previous

Explanation:

Step1: Identify the function type

The height function \(h(t)=-16t^{2}+90t\) is a quadratic function in the form \(y = ax^{2}+bx + c\) (here \(c = 0\)), where \(a=-16\), \(b = 90\). For a quadratic function \(y=ax^{2}+bx + c\), the vertex of the parabola (which gives the maximum value when \(a<0\)) has its \(t\) - coordinate at \(t=-\frac{b}{2a}\).

Step2: Calculate the time \(t\) at which maximum height occurs

Substitute \(a=-16\) and \(b = 90\) into the formula \(t=-\frac{b}{2a}\).

$$t=-\frac{90}{2\times(-16)}=\frac{90}{32}=\frac{45}{16}=2.8125$$

Step3: Calculate the maximum height

Substitute \(t = 2.8125\) into the height function \(h(t)=-16t^{2}+90t\).

$$h(2.8125)=-16\times(2.8125)^{2}+90\times(2.8125)$$

First, calculate \((2.8125)^{2}=7.91015625\)
Then, \(-16\times7.91015625=-126.5625\)
And \(90\times2.8125 = 253.125\)

$$h(2.8125)=- 126.5625+253.125=126.5625\approx126.56$$

Answer:

\(126.56\)