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Antonio threw a ball with an upward velocity of 6 meters per second from a height of 8 meters. The formula below describes this situation. Which is closest to the time it will take the ball to hit the ground? h(x) = – 4.9t^2 + 60t + 8

User Nowaq
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1 Answer

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Antonio threw a ball with an upward velocity of 6 meters per second from a height of 8 meters. The formula below describes this situation. Which is closest to the time it will take the ball to hit the ground? h(x) = – 4.9t^2 + 60t + 8​

we know that

when the ball hit the ground, the function h(x) is equal to zero

so

– 4.9t^2 + 6t + 8=0

using the quadratic formula


t=\frac{-b\pm\sqrt[\square]{b^2-4ac}_{}}{2a}

we have

a=-4.9

b=6

c=8

substitute the values in the formula


\begin{gathered} t=\frac{4.9\pm\sqrt[\square]{(-4.9)^2-4(-4.9)(8)}_{}}{2(-4.9)} \\ \\ t=\frac{4.9\pm\sqrt[\square]{180.81}_{}}{-9.8} \\ \\ t=\frac{4.9\pm13.45_{}}{-9.8} \\ \\ t1==\frac{4.9+13.45_{}}{-9.8}=-1.87\text{ -}\longrightarrow\text{ the time can not be negative} \\ t2=\frac{4.9-13.45_{}}{-9.8}=0.87\text{ sec} \end{gathered}

therefore

the answer is

0.87 seconds

User BeeFriedman
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