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A ball is thrown from an initial height of 2 meters with an initial upward velocity of 5 m/s.

h=2+5t-5t^2
The ball's height of (in meters) after seconds is given by the following. Find all values of for which the ball's height is 3 meters.

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Answer: The height of the ball (in meters) after t seconds is given by the equation h = 2 + 5t - 5t^2. We can use this equation to find the values of t for which the ball's height is 3 meters.

We need to find the values of t that make h = 3, so we can substitute 3 for h in the equation:

3 = 2 + 5t - 5t^2

Then we can solve for t by moving all the terms to one side of the equation and factoring it:

-5t^2 + 5t - 1 = 0

And we can use the quadratic formula to find the solutions of this equation:

t = (5 +/- sqrt(5^2 - 4*(-1)(-1)))/ 2(-1)

t = (5 +/- sqrt(25 + 4))/ -2

t = (5 +/- sqrt(29))/ -2

The solutions of this equation are t = (5 +/- sqrt(29))/ -2

Since the time, t, is a real value the square root of 29 should be positive.

The values of t for which the ball's height is 3 meters are not real values, so there's no time when the ball's height is 3 meters.

Explanation:

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