h=5+18t-16t^2

Question

You are hitting baseballs. When you toss the ball into the air, your hand is 5 feet above the ground. You hit the ball when it falls back to a height of 3.5 feet. You toss the ball with an initial velocity of 18 feet per second. The height of the ball seconds after leaving your hand is given by h=5+18t-16t^2. About how much time passes before you hit the ball?

Answer 1

"You toss the ball with an initial velocity of 18 feet per second. The height of the ball t seconds after leaving your hand is given by h=5+18t-16t^2."

 

The height (when batted) is 3.5, so solve this quadratic equation: 3.5 = 5+18t-16t²  -- accept only a positive t value.