a ball is thrown from a height of 64 meters with an initial downward velocity of 3 m/s. the balls height h…

a ball is thrown from a height of 64 meters with an initial downward velocity of 3 m/s. the balls height h (in meters) after t seconds is given by the following.\nh = 64 - 3t - 5t^2\nhow long after the ball is thrown does it hit the ground?\nround your answer(s) to the nearest hundredth.\n(if there is more than one answer, use the \or\ button.)

a ball is thrown from a height of 64 meters with an initial downward velocity of 3 m/s. the balls height h (in meters) after t seconds is given by the following.\nh = 64 - 3t - 5t^2\nhow long after the ball is thrown does it hit the ground?\nround your answer(s) to the nearest hundredth.\n(if there is more than one answer, use the \or\ button.)

Answer

Explanation:

Step1: Set height to 0

When the ball hits the ground, $h = 0$. So we set up the equation $0=64 - 3t-5t^{2}$.

Step2: Rearrange to standard quadratic form

We rewrite the equation as $5t^{2}+3t - 64=0$.

Step3: Use quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 5$, $b = 3$, and $c=-64$. First, calculate the discriminant $\Delta=b^{2}-4ac=(3)^{2}-4\times5\times(-64)=9 + 1280=1289$.

Step4: Find t - values

$t=\frac{-3\pm\sqrt{1289}}{10}$. We have two solutions for $t$: $t_1=\frac{-3+\sqrt{1289}}{10}$ and $t_2=\frac{-3 - \sqrt{1289}}{10}$. Since time cannot be negative in this context, we discard $t_2$. Calculate $t_1=\frac{-3+\sqrt{1289}}{10}\approx\frac{-3 + 35.9}{10}=\frac{32.9}{10}=3.29$.

Answer:

$t\approx3.29$ seconds