Question

when a baseball is hit by a batter, the height of the ball, h(t), at time t, t ≥ 0, is determined by the equation h(t) = -16t² + 64t + 4. if t is in seconds, for which interval of time is the height of the ball greater than or equal to 52 feet?
○ 1 < t < 3
○ t ≤ 1
○ 0 < t < 1
○ 1 ≤ t ≤ 3

Explanation:

Step1: Set up the inequality

We need to find when \( h(t) \geq 52 \). Given \( h(t)= - 16t^{2}+64t + 4 \), so we set up the inequality:

$$ -16t^{2}+64t + 4\geq52 $$

Subtract 52 from both sides:

$$ -16t^{2}+64t + 4-52\geq0 $$

Simplify to get:

$$ -16t^{2}+64t - 48\geq0 $$

Multiply both sides by - 1 (and reverse the inequality sign):

$$ 16t^{2}-64t + 48\leq0 $$

Divide both sides by 16:

$$ t^{2}-4t + 3\leq0 $$

Step2: Solve the quadratic equation

Factor the quadratic \( t^{2}-4t + 3 \). We need two numbers that multiply to 3 and add to - 4. The numbers are - 1 and - 3. So,

$$ t^{2}-4t + 3=(t - 1)(t - 3) $$

Set \( (t - 1)(t - 3)=0 \). The solutions are \( t = 1 \) and \( t = 3 \).

Step3: Analyze the inequality

The quadratic function \( y=t^{2}-4t + 3 \) is a parabola opening upwards (since the coefficient of \( t^{2} \) is positive). The inequality \( t^{2}-4t + 3\leq0 \) holds for the values of \( t \) between the roots (including the roots). So the solution to \( t^{2}-4t + 3\leq0 \) is \( 1\leq t\leq3 \).

Answer:

\( 1\leq t\leq3 \) (the option with "1 ≤ t ≤ 3")