A ball is thrown from an initial height of 3 meters with an initial upward velocity of 20 m/s The ball's height h (in meters) after t seconds is given by the following.
h=3+20t-5t^2
Find all values of t for which the ball's height is 13 meters.
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
To find the values of t for which the ball's height is 13 meters, we will set h = 13 in the equation and solve for t.
h = 3 + 20t - 5t^2
13 = 3 + 20t - 5t^2
10 = 20t - 5t^2
5t^2 - 20t + 10 = 0
Now, we will solve this quadratic equation using the quadratic formula:
t = (-(-20) ± √((-20)^2 - 4*5*10)) / 2*5
t = (20 ± √(400 - 200)) / 10
t = (20 ± √200) / 10
t = (20 ± 14.14) / 10
This gives us two possible values for t:
t = (20 + 14.14) / 10 ≈ 3.41
t = (20 - 14.14) / 10 ≈ 0.59
Therefore, the ball's height is 13 meters at t ≈ 3.41 seconds or t ≈ 0.59 seconds.