Answer: The ball hits the ground after 4.315 s
Step-by-step explanation:
We are given the vertical motion equation of the ball in the form [tex]y=y_{o}+V_{o}t+\frac{1}{2}gt^{2}[/tex] as follows:
[tex]h=-50-10t+5t^{2}[/tex]
If we want to find the time [tex]t[/tex] in which the ball hits the ground ([tex]h=0[/tex]), we have to rewrite the equation as shown below:
[tex]0=-50-10t+5t^{2}[/tex]
Aplying common factor [tex]5[/tex] in the right side of the equation:
[tex]0=5(-10-2t+t^{2})[/tex]
Rearranging the equation:
[tex]0=t^{2}-2t-10[/tex]
At this point we have a quadratic equation of the form [tex]0=at^{2}+bt+c[/tex], and we have to use the quadratic formula if we want to find [tex]t[/tex]:
[tex]t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]
Where [tex]a=1[/tex], [tex]b=-2[/tex], [tex]c=-10[/tex]
Substituting the known values and choosing the positive result of the equation:
[tex]t=\frac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(-10)}}{2(1)}[/tex]
[tex]t=4.315 s[/tex] This is the time it takes to the ball to hit the ground
