Question

a ball is thrown from an initial height of 6 feet with an initial upward velocity of 33 ft/s. the balls height ( h ) (in feet) after ( t ) seconds is given by the following.
( h = 6 + 33t - 16t^2 )
find all values of ( t ) for which the balls height is 22 feet.
round your answer(s) to the nearest hundredth.
(if there is more than one answer, use the \or\ button.)

Explanation:

Step1: Set height to 22 feet

$$22 = 6 + 33t - 16t^2$$

Step2: Rearrange to standard quadratic form

$$16t^2 - 33t + 16 = 0$$

Step3: Apply quadratic formula $t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, $a=16$, $b=-33$, $c=16$:
$$t=\frac{33\pm\sqrt{(-33)^2-4(16)(16)}}{2(16)}$$

Step4: Calculate discriminant

$$\sqrt{1089 - 1024}=\sqrt{65}\approx8.06$$

Step5: Compute two t values

$$t_1=\frac{33+8.06}{32}\approx\frac{41.06}{32}\approx1.28$$
$$t_2=\frac{33-8.06}{32}\approx\frac{24.94}{32}\approx0.78$$

Answer:

$t \approx 0.78$ seconds or $t \approx 1.28$ seconds