The height $h$ above the ground level reached by a ball , $t$ seconds after it is thrown is given by $h (t) = - 16 t^{2} + 46 t + 5$ . Calculate the approximate maximum height the ball reaches. (All lengths in feet)

5

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23

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35

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38

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46

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Solution

The correct option is D $38$

The quadratic equation of standard form is
$y = a (x - h)^{2} + k$
So use the method of completing the square to rewrite the equation in vertex form.

Then, read the value of $k$ to find the maximum height of the projectile.

$h = - 16 t^{2} + 46 t + 5$

$h = - 16 (t^{2} - 2.875 t + 2.066) + 5 - (- 16 \times 2.066)$

$h = - 16 (t - 2)^{2} + 5 - (- 33.056)$

$h = - 16 (t - 1.4375)^{2} + 38.056$

The vertex is $(1.4, 38)$ , therefore, the maximum height is $38$ feet.

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The height h above the ground level reached by a ball , t seconds after it is thrown is given by h(t)=−16t2+46t+5 . Calculate the approximate maximum height the ball reaches. (All lengths in feet)

The height $h$ above the ground level reached by a ball , $t$ seconds after it is thrown is given by $h (t) = - 16 t^{2} + 46 t + 5$ . Calculate the approximate maximum height the ball reaches. (All lengths in feet)