Question

a baseball is thrown upwards from a height of 5 feet with an initial speed of 64 feet per second, and its height h (in feet) from the ground is given by h(t) = 5 + 64t - 16t² where t is time in seconds. using a graphing calculator, determine at what time the ball reaches its maximum height.
a. 0 seconds
b. 1 second
c. 2 seconds
d. 3 seconds

Explanation:

Step1: Recognize the function type

The height function \( h(t) = 5 + 64t - 16t^2 \) is a quadratic function in the form \( h(t)=at^2 + bt + c \), where \( a=-16 \), \( b = 64 \), \( c = 5 \). For a quadratic function, the vertex (which gives the maximum or minimum value) occurs at \( t=-\frac{b}{2a} \) when \( a<0 \) (since the parabola opens downwards, the vertex is the maximum point).

Step2: Calculate the time of maximum height

Substitute \( a=-16 \) and \( b = 64 \) into the formula \( t=-\frac{b}{2a} \).

$$ t=-\frac{64}{2\times(-16)}=-\frac{64}{-32} = 2 $$

We can also verify this by analyzing the options. At \( t = 0 \), \( h(0)=5+0 - 0=5 \) (initial height). At \( t = 1 \), \( h(1)=5 + 64\times1-16\times1^2=5 + 64 - 16 = 53 \). At \( t = 2 \), \( h(2)=5+64\times2 - 16\times2^2=5 + 128 - 64 = 69 \). At \( t = 3 \), \( h(3)=5+64\times3-16\times3^2=5 + 192 - 144 = 53 \). So the height is maximum at \( t = 2 \) seconds.

Answer:

C. 2 seconds