Question

1. the height, h, over time, t, of a rock thrown from a height of 48 feet with an initial velocity of 32 feet/second is modeled by the equation h = -16t² + 32t + 48. assuming the rock stops when it hits the ground, what are the constraints on the variable t? \\(\square < t < \square\\) number pad with 0-9, +, -, ×, ÷, =, arrow keys, and x button

Explanation:

Step1: Set \( h = 0 \)

When the rock hits the ground, \( h = 0 \). So we have the equation \( - 16t^{2}+32t + 48=0 \).

Step2: Divide the equation by \(-16\)

Divide each term by \(-16\) to simplify: \( t^{2}-2t - 3=0 \).

Step3: Factor the quadratic equation

Factor \( t^{2}-2t - 3=(t - 3)(t + 1)=0 \).

Step4: Solve for \( t \)

Set each factor equal to zero: \( t-3 = 0\) gives \( t = 3\); \( t + 1=0\) gives \( t=-1\). Since time \( t\geq0\), we discard \( t=-1\).

Answer:

\( t\geq0\) and \( t = 3\) is the time when it hits the ground, so the constraint on \( t\) is \(0\leq t\leq3\)