Question

a ball is thrown from a height of 205 feet with an initial downward velocity of 11 ft/s. the balls height h (in feet) after t seconds is given by the following.
h = 205 - 11t - 16t^2
how long after the ball is thrown does it hit the ground?
round your answer(s) to the nearest hundredth.
(if there is more than one answer, use the \or\ button.)

Explanation:

Step1: Set height to 0

When the ball hits the ground, $h = 0$. So we set up the equation $0=205 - 11t-16t^{2}$. Rearrange it to the standard quadratic - form $16t^{2}+11t - 205 = 0$.

Step2: Use the quadratic formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 16$, $b = 11$, and $c=-205$. First, calculate the discriminant $\Delta=b^{2}-4ac=(11)^{2}-4\times16\times(-205)=121 + 13120=13241$.

Step3: Find the values of t

$t=\frac{-11\pm\sqrt{13241}}{2\times16}=\frac{-11\pm115.07}{32}$. We have two solutions: $t_1=\frac{-11 + 115.07}{32}=\frac{104.07}{32}\approx3.25$ and $t_2=\frac{-11 - 115.07}{32}=\frac{-126.07}{32}\approx - 3.94$. Since time cannot be negative in this context, we discard the negative solution.

Answer:

$t\approx3.25$ seconds