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.
To find when the ball is 6 meters above the ground, we need to find the value of t that satisfies the equation h = 6.
The equation given is h = -5t^2 + 10t + 2. Setting this equal to 6, we have:
-5t^2 + 10t + 2 = 6
Rearranging the equation:
-5t^2 + 10t - 4 = 0
To solve this quadratic equation for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -5, b = 10, and c = -4. Substituting these values into the formula:
t = (-10 ± √(10^2 - 4(-5)(-4))) / (2(-5))
Simplifying further:
t = (-10 ± √(100 - 80)) / (-10)
t = (-10 ± √20) / (-10)
This gives us two possible solutions for t, found by substituting the positive and negative values of the square root:
t1 = (-10 + √20) / (-10)
t2 = (-10 - √20) / (-10)
To find the values of t to the nearest tenth of a second, you can use a calculator or do the calculations manually:
Calculating t1:
t1 = (-10 + √20) / (-10) ≈ 0.3 seconds (rounded to the nearest tenth)
Calculating t2:
t2 = (-10 - √20) / (-10) ≈ -1.3 seconds (rounded to the nearest tenth)
The reason why there are two answers is because the ball reaches the height of 6 meters twice during its trajectory - once on the way up and once on the way down. The negative solution, t2, represents the time when the ball is at the same height during the descent. However, in the context of the problem, we usually consider only positive time values, so the approximate answer is t ≈ 0.3 seconds.