A ball is thrown straight up. Its height, h(in metres), after t seconds is given by h=-5t^2+10t+2. To the nearest tenth of a second, when is the ball 6m above the ground?
Explain why there are two answers.
I don't get how to calculate it. Please help.
whats the question
when is the ball 6m above the ground?
To the nearest tenth of a second,
To find the time when the ball is 6 meters above the ground, we need to solve the equation h = 6, where h is the height of the ball. Given the equation h = -5t^2 + 10t + 2, we substitute 6 for h:
6 = -5t^2 + 10t + 2
Rearranging the equation, we get:
5t^2 - 10t - 4 = 0
To solve this quadratic equation, we can either use factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula because factoring is not easily apparent:
t = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 5, b = -10, and c = -4. Substituting these values into the quadratic formula, we get:
t = (-(-10) ± √((-10)^2 - 4(5)(-4))) / (2(5))
= (10 ± √(100 + 80)) / 10
= (10 ± √180) / 10
Simplifying further, we have:
t = (10 ± √(2 * 2 * 3 * 5)) / 10
= (10 ± 2√(5 * 3)) / 10
= (10 ± 2√15) / 10
Now, we have two possible values for t:
1. t = (10 + 2√15) / 10
2. t = (10 - 2√15) / 10
To the nearest tenth of a second, we can find the decimal approximation for each of these values:
1. t ≈ (10 + 2(3.87)) / 10 ≈ 1.774
2. t ≈ (10 - 2(3.87)) / 10 ≈ 0.226
Therefore, the ball is approximately 6 meters above the ground at t ≈ 1.8 seconds and t ≈ 0.2 seconds.
Now, let's explain why there are two answers. In this case, we have a quadratic equation, which typically has two solutions. In the context of the problem, a ball is thrown straight up, reaches its maximum height, and then falls back down. Thus, the ball passes the 6-meter mark twice, once on the way up and once on the way down. These two points in time correspond to the two solutions of the quadratic equation.