Question

a model rocket is launched with an initial upward velocity of 100 ft/s. the rockets height (h) (in feet) after (t) seconds is given by the following.(h = 100t-16t^{2})find all values of (t) for which the rockets height is 34 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 $h = 34$ in the equation $h=100t - 16t^{2}$, so we get $34 = 100t-16t^{2}$. Rearrange it to the standard - form of a quadratic equation $16t^{2}-100t + 34 = 0$. Divide through by 2 to simplify: $8t^{2}-50t + 17 = 0$.

Step2: Apply the quadratic formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For the equation $8t^{2}-50t + 17 = 0$, $a = 8$, $b=-50$, and $c = 17$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-50)^{2}-4\times8\times17=2500 - 544 = 1956$.

Step3: Calculate the values of t

$t=\frac{50\pm\sqrt{1956}}{16}$. Since $\sqrt{1956}\approx44.23$, then $t=\frac{50\pm44.23}{16}$.

• For the plus - sign: $t_1=\frac{50 + 44.23}{16}=\frac{94.23}{16}\approx5.89$.
• For the minus - sign: $t_2=\frac{50 - 44.23}{16}=\frac{5.77}{16}\approx0.36$.

Answer:

$t\approx0.36$ or $t\approx5.89$