Respuesta :
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
[tex]t=\frac{-b\pm\sqrt[\square]{b^2-4ac}_{}}{2a}[/tex]we have
a=-4.9
b=6
c=8
substitute the values in the formula
[tex]\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}[/tex]therefore
the answer is
0.87 seconds
