Question

a ball is thrown from a height of 25 meters with an initial downward velocity of 5 m/s. the balls height h (in meters) after t seconds is given by the following:
$h = 25 - 5t - 5t^2$
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.)

t = \square seconds

Explanation:

Step1: Set h = 0 (ground level)

We need to solve the equation \(0 = 25 - 5t - 5t^2\). First, we can rewrite the quadratic equation in standard form \(ax^2+bx + c = 0\). So, \(5t^2+ 5t - 25=0\) (multiplying both sides by - 1 to make the coefficient of \(t^2\) positive). We can simplify this equation by dividing all terms by 5: \(t^2 + t - 5=0\).

Step2: Use 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 \(t^2 + t - 5=0\), we have \(a = 1\), \(b = 1\), and \(c=- 5\).

First, calculate the discriminant \(\Delta=b^2-4ac=(1)^2-4\times1\times(-5)=1 + 20=21\).

Then, \(t=\frac{-1\pm\sqrt{21}}{2}\).

We have two solutions:
\(t_1=\frac{-1+\sqrt{21}}{2}\) and \(t_2=\frac{-1 - \sqrt{21}}{2}\)

Since time cannot be negative, we discard \(t_2\) (because \(\sqrt{21}\approx4.583\), so \(\frac{-1 - 4.583}{2}\approx-2.791\) which is negative).

Now, calculate \(t_1=\frac{-1+\sqrt{21}}{2}\approx\frac{-1 + 4.583}{2}=\frac{3.583}{2}\approx1.79\)

Answer:

\(t\approx1.79\) seconds