Question

question 3 of 14
a ball is thrown from a height of 67 meters with an initial downward velocity of 5 m/s. the balls height h (in meters) after t seconds is given by the following.
h = 67 - 5t - 5t^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 $0=67 - 5t-5t^{2}$. Rearrange it to the standard quadratic - form $5t^{2}+5t - 67 = 0$.

Step2: Identify coefficients

For the quadratic equation $ax^{2}+bx + c = 0$ (in our case, $x=t$, $a = 5$, $b = 5$, and $c=-67$).

Step3: Apply quadratic formula

The quadratic formula is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute the values: $t=\frac{-5\pm\sqrt{5^{2}-4\times5\times(-67)}}{2\times5}=\frac{-5\pm\sqrt{25 + 1340}}{10}=\frac{-5\pm\sqrt{1365}}{10}$.

Step4: Calculate the two solutions

$\sqrt{1365}\approx36.946$. Then $t_{1}=\frac{-5 + 36.946}{10}=\frac{31.946}{10}=3.1946$ and $t_{2}=\frac{-5 - 36.946}{10}=\frac{-41.946}{10}=-4.1946$.

Step5: Reject the non - physical solution

Since time $t\geq0$, we reject the negative solution.

Answer:

$t\approx3.19$