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.

Question

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.)

Answer 1

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.