Question

problem 11. a ball is thrown in the air. the ball’s height is described by the quadratic equation $h(t) = -t^2 + 26t$, where $t$ is in seconds and $h$ is inches. this question has three parts: (a) when does the ball reach its maximum height? (b) what is the maximum height of the ball? (c) when does the ball return to the person who threw it’s hand?

Explanation:

Part (a)

Step1: Recall vertex formula for parabola

For a quadratic function \( H(t) = at^2 + bt + c \), the time at which the vertex (maximum or minimum) occurs is given by \( t = -\frac{b}{2a} \). Here, \( a = -1 \), \( b = 26 \), \( c = 0 \).

Step2: Calculate time for max height

Substitute \( a = -1 \) and \( b = 26 \) into the formula: \( t = -\frac{26}{2\times(-1)} = \frac{-26}{-2} = 13 \).

Step1: Substitute t = 13 into H(t)

We know from part (a) that the time of maximum height is \( t = 13 \). Substitute \( t = 13 \) into \( H(t)=-t^2 + 26t \).

Step2: Calculate H(13)

\( H(13)=-(13)^2 + 26\times13=-169 + 338 = 169 \).

Step1: Set H(t) = 0

The ball returns to the hand when its height \( H(t) = 0 \). So we solve the equation \( -t^2 + 26t = 0 \).

Step2: Factor the equation

Factor out \( t \): \( t(-t + 26)=0 \). This gives two solutions: \( t = 0 \) (when it is thrown) and \( -t + 26 = 0 \Rightarrow t = 26 \).

Answer:

The ball reaches its maximum height at \( t = 13 \) seconds.

Part (b)