Question

1. free-falling object you throw a baseball upward with an initial velocity of 30 feet per second. the height ( h ) (in feet) of the baseball relative to your glove is modeled by the position function ( h(t)=-16 t^{2}+30 t ), where ( t ) is the time (in seconds). how long does it take for the ball to reach your glove?

Explanation:

Step1: Set up the equation

We want to find when \(h(t) = 32\). So we set up the equation \(h(t)=-16t^{2}+32t = 32\).

Step2: Rearrange the equation

Rearrange it to the standard quadratic form \(ax^{2}+bx + c=0\). We get \(16t^{2}-32t + 32=0\). Divide through by \(16\) to simplify: \(t^{2}-2t + 2=0\).

Step3: Use the quadratic formula

The quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For the equation \(t^{2}-2t + 2=0\), where \(a = 1\), \(b=-2\), and \(c = 2\). First, calculate the discriminant \(\Delta=b^{2}-4ac=(-2)^{2}-4\times1\times2=4 - 8=-4\).

Step4: Analyze the discriminant

Since \(\Delta=-4<0\), there are no real - valued solutions. But wait, we made a mistake in step 2. Let's start over.

We want \(h(t)=-16t^{2}+32t=32\). Rearrange correctly: \(-16t^{2}+32t-32 = 0\). Divide by \(-16\): \(t^{2}-2t + 2=0\) (wrong). Wait, correct rearrangement: \(h(t)=-16t^{2}+32t\). Set \(h(t) = 0\) (height relative to glove, when it comes back to the glove). \( - 16t^{2}+32t=0\). Factor out \(-16t\): \(-16t(t - 2)=0\).

Step5: Solve for \(t\)

Set each factor equal to zero. \(-16t=0\) gives \(t = 0\) (initial time when thrown). \(t-2=0\) gives \(t = 2\) seconds.

Answer:

\(t = 2\) seconds.