Well, it's a quadratic, so it has to have 2 roots. 'Roots' meaning when its velocity is 0 or in this case, the ball hits the ground. So, let's just throw this into the quadratic equation. Remember, the quadratic equation is:
[tex]x = \frac{-b + \sqrt{b^2-4ac} }{2a} [/tex]
and
[tex]x = \frac{-b - \sqrt{b^2-4ac} }{2a} [/tex]
To plug values in, let's first rearrange the equation, so it's easier to plug in:
[tex]h = -5t^2-10t+65[/tex]
Plug in (a = -5, b = -10, c = 65) and since we are solving for t, not x:
[tex]0 = -5t^2-10t+65[/tex]
[tex]t = \frac{10+ \sqrt{(-10)^2-4(-5)(65)} }{2(-5)} [/tex]
and
[tex]t = \frac{10+ \sqrt{(-10)^2-4(-5)(65)} }{2(-5)} [/tex]
Solve:
[tex]t = \frac{10+ \sqrt{100+1300} }{-10} = \frac{10+ \sqrt{1400} }{-10} [/tex]
and
[tex]t = \frac{10- \sqrt{100+1300} }{-10} = \frac{10- \sqrt{1400} }{-10}[/tex]
Simplify:
[tex]t = \frac{10+37.42}{-10} =-4.742[/tex]
and
[tex]t = \frac{10-37.42}{-10} = 2.742[/tex]
Since time cannot be negative, your answer is the ball takes 2.742 seconds to hit the ground.
