Question

a ball is thrown from a height of 32 meters with an initial downward velocity of 2 m/s. the balls height h (in meters) after t seconds is given by the following.
$h = 32 - 2t - 5t^{2}$
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 up the equation

When the ball hits the ground, \(h = 0\). So we set \(32-2t - 5t^{2}=0\). Rearranging gives \(5t^{2}+2t - 32=0\).

Step2: Use the quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 5\), \(b = 2\), \(c=-32\)), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Substitute the values: \(t=\frac{-2\pm\sqrt{2^{2}-4\times5\times(-32)}}{2\times5}=\frac{-2\pm\sqrt{4 + 640}}{10}=\frac{-2\pm\sqrt{644}}{10}\).

Step3: Calculate the roots

\(\sqrt{644}\approx25.38\).
\(t_{1}=\frac{-2 + 25.38}{10}=\frac{23.38}{10}=2.34\) (approx).
\(t_{2}=\frac{-2-25.38}{10}=\frac{-27.38}{10}=-2.74\) (discard since time \(t\geq0\)).

Answer:

\(t = 2.34\) seconds