Answer:
[tex]1+2\sqrt{2}[/tex] seconds
Step-by-step explanation:
When the ball hits the ground, its height will obviously be 0. Therefore, you can set up the equation the following way:
[tex]0=-5t^2+10t+35 \\\\-5(t^2-2t-7)=0[/tex]
Plugging this into the quadratic equation, you get:
[tex]t=\dfrac{2\pm\sqrt{4+28}}{2}= \\\\\dfrac{2\pm4\sqrt{2}}{2}= \\\\1\pm2\sqrt{2}[/tex]
Since the time must be positive, it takes the ball [tex]1+2\sqrt{2}[/tex] seconds to hit the ground, or around 3.828. Hope this helps!
