Question

a ball is thrown from an initial height of 7 feet with an initial upward velocity of 40 ft/s. the balls height ( h ) (in feet) after ( t ) seconds is given by the following. ( h = 7+40t - 16t^{2}) find all values of ( t ) for which the balls height is 29 feet. round your answer(s) to the nearest hundredth. (if there is more than one answer, use the \or\ button.)

Explanation:

Step1: Set up the equation

Set $h = 29$ in the equation $h=7 + 40t-16t^{2}$. So we get $29=7 + 40t-16t^{2}$.

Step2: Rearrange to standard quadratic form

Rearrange the equation to $16t^{2}-40t + 22 = 0$. Divide through by 2 to simplify: $8t^{2}-20t + 11 = 0$.

Step3: Apply quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 8$, $b=-20$, $c = 11$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-20)^{2}-4\times8\times11=400 - 352 = 48$.

Step4: Find the values of t

$t=\frac{20\pm\sqrt{48}}{16}=\frac{20\pm4\sqrt{3}}{16}=\frac{5\pm\sqrt{3}}{4}$.
$t_1=\frac{5+\sqrt{3}}{4}\approx\frac{5 + 1.732}{4}=\frac{6.732}{4}=1.683\approx1.68$.
$t_2=\frac{5-\sqrt{3}}{4}\approx\frac{5 - 1.732}{4}=\frac{3.268}{4}=0.817\approx0.82$.

Answer:

$t = 0.82$ or $t = 1.68$