Answer:
Step-by-step explanation:
[tex]h(t)=-4.9t^2+24t+2[/tex] is our quadratic,where h(t) is the height of the object after t seconds. If we are looking for how long, t, it takes the object to hit the ground, h(t) will be equal to 0 since the height of something on the ground is 0. We set the quadratic equal to zero and factor it using the quadratic formula:
[tex]t=\frac{-24+/-\sqrt{24^2-4(-4.9)(2)} }{2(-4.9)} }[/tex] and
[tex]t=\frac{-24+/-\sqrt{576+39.2} }{-9.8}[/tex] and
[tex]t=\frac{-24+/-\sqrt{615.2} }{-9.8}[/tex] and
the 2 solutions are:
[tex]t=\frac{-24+24.803}{-9.8}=-.08 sec[/tex] and
[tex]t=\frac{-24-24.803}{-9.8}=4.98sec[/tex]
and since time can never be negative, we know that it takes 4.98 seconds for the rock to hit the ground under the given conditions.
