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A ball is thrown from an initial height of 3 feet with an initial upward velocity of 29 s. The ball's heighth (in feet) after seconds to given by the following:H=3+29t-16t^2Find all values of for which the ball's height is 15 feetRound your answer(s) to the nearest hundredth(if there is more than one answer, use the "of" button)

Answer :

The height of the ball at time t is modeled by

[tex]H(t)=3+29t-16t^2[/tex]

To find the values of t for which the height is 15 feet, let us solve the equation

[tex]\begin{gathered} H(t)=15 \\ 3+29t-16t^2=15 \\ 16t^2-29t+12=0 \\ x=\frac{29\pm\sqrt[]{73}}{32} \\ =0.64,1.17 \end{gathered}[/tex]

So, at t=0.6 sec and =1.173 sec, the height of the ball is 15 feet (Rounding off to the nearest hundredth)

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