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 \( h = 0 \)

When the ball hits the ground, \( h = 0 \). So we have the equation \( 0=32 - 2t-5t^{2}\). Rearranging it to the standard quadratic form \( ax^{2}+bx + c = 0 \), we get \( 5t^{2}+2t - 32=0\), where \( a = 5\), \( b = 2\), and \( c=-32\).

Step2: Use the quadratic formula \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \)

Substitute \( a = 5\), \( b = 2\), and \( c=-32\) into the formula:

$$
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$$

We have two solutions: \( t_{1}=\frac{-1+\sqrt{161}}{5}\) and \( t_{2}=\frac{-1 - \sqrt{161}}{5}\). Since time \( t>0\), we discard \( t_{2}\) (because \(\frac{-1-\sqrt{161}}{5}<0\)).

Step3: Calculate the value of \( t_{1}\)

\(\sqrt{161}\approx12.69\), then \( t=\frac{-1 + 12.69}{5}=\frac{11.69}{5}=2.34\)

Answer:

\(t = 2.34\) seconds