A ball is thrown from an initial height of 3 meters with an initial upward velocity of 20m/s. The ball's height (in meters) after seconds is given by the following h=3+20t-5t^2. Find all values of for which the ball's height is 13 meters.
+37159 h = 13 meters
13 = -5t^2 + 20t + 3
-5t^2 + 20t -10 = 0 we can simplify this a bit by dividing through by -5
t^2 -4t +2 = 0 Now use the quadratic formula to find the t values (two of them) ...one will be for the ball on the way UP
the larger one will be for whe the ball is coming DOWN .
a = 1 b= -4 c = 2
\(t = {-(-4) \pm \sqrt{(-4)^2-4(1)(2)} \over 2(1)}\) I will let you finish.....
