A ball is thrown from an initial height of 3 meters with an initial upward velocity of 20 m/s The ball's height (in meters) after t seconds is given by the following: h= 3+20t-5t^2

Question

Find all values of for which the ball's height is meters.

Answer 1

(0.268, 8) and (3.732, 8)

Thanks Nathan for clarifying things in the comments.

t = 0.268 seconds

t= 3.32 seconds

Find all values of t for which the ball's height is 8 meters.

h = -5t² + 20t + 3

8 = -5t² + 20t + 3

Subtract 8 from both sides of the equation

0 = -5t² + 20t - 5

This is a parabola which opens downward

a = -5

b = 20

c = -5

the vertex -b/2a, h; (2, 23)

-20/2(-5) = -20/-10 = 2

h = -5(2²) + 20(2) + 3

h = -20 + 40 + 3 = 23

Use the Quadratic Formula to solve for t

0 = -5t² + 20t - 5

Dividing both sides of the equation by 5 makes the math easier

0 = -t² + 4t -1

Now

a = -1

b = 4

c = 1

(-b±(√b²-4ac))/2a

(-4(±√16 - 4))/-2

(-4±(√12))/-2

-4/-2 + 2√3/-2

2 - √3 = 0.26795 seconds

-4/-2 - 2√3/-2

2 + √3 = 3.732 seconds

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