st-math1-q29

The function $h$ given by $h (t) = - {16 t}^{2} + 46 t + 5$ represents the height of a ball, in feet, $t$ seconds after it is thrown. To the nearest foot, what is the maximum height the ball reaches?

Select an Answer

$5$

Correct Answer:

$23$

Correct Answer:

$35$

Correct Answer:

$38$

Correct Answer:

Yes

$46$

Correct Answer:

View Correct Answer

The correct answer is D.

One way to determine the maximum height the ball reaches is to rewrite the quadratic expression that defines the function $h$ by completing the square:

$table row cell negative 16 t squared plus 46 t plus 5 end cell equals cell negative 16 open parentheses t squared minus 23 over 8 t close parentheses plus 5 end cell end table space space space space space space space space space space space space space space space space space space space space space space space space space table row equals cell negative 16 open parentheses t squared minus 23 over 8 t plus open parentheses 23 over 16 close parentheses squared minus open parentheses 23 over 16 close parentheses squared close parentheses plus 5 end cell end table table row cell space space space space space space space space space space space space space space space space space space space space space space space end cell equals cell negative 16 end cell end table open parentheses open parentheses t minus 23 over 16 close parentheses squared minus open parentheses 23 over 16 close parentheses squared close parentheses plus 5 table row cell space space space space space space space space space space space space space space space space space space space space space space space end cell equals cell negative 16 open parentheses t minus 23 over 16 close parentheses squared plus 16 open parentheses 23 over 16 close parentheses squared plus 5 end cell end table$

It is not necessary to simplify any further, as the maximum height must correspond to $t = \frac{23}{16}$ , which is the only value of $t$ that makes the term $- 16 {(t - \frac{23}{16})}^{2}$ nonnegative. By substitution, $h (\frac{23}{16}) = 16 {(\frac{23}{16})}^{2} + 5 = 38.0625$ . Therefore, to the nearest foot, the maximum height the ball reaches is $38$ feet.

Alternatively, one can use a graphing calculator to determine the maximum value of the function $h$ . Since $h (0) = 5,$ $h (1) = - 16 + 46 + 5 = 35$ , $h (2) = - 16 (4) + 46 (2) + 5 = 33$ and $h (3) = - 16 (9) + 46 (3) + 5 = - 1$ , the maximum value of $h$ must occur for some $t - value$ between $1$ and $2$ . Set the window so that the independent variable goes from $0$ to $3$ and the dependent variable goes from $0$ to $50$ to view the vertex of the parabola. Upon tracing the graph, the maximum value of $h$ is slightly greater than $38$ . Therefore, to the nearest foot, the maximum height the ball reaches is $38$ feet.