A ball is thrown straight up. Its height, h (in metres), after t seconds is given by h = -5t2+10t+2. To the nearest tenth of a second, when is the ball 6m above the ground? Explain why there are two answers.
in depth answer please!
Enter powers this way : -5^2
You want h to be 6, so
-5t^2+10t+2 = 6
5t^2 - 10t + 4 = 0
use formula since it doesn't factor
t = (10 ± √(100 -4(-5)(4)) )/10
= (10 ± √180)/10
= appr 2.34 or -.34
the corresponding parabola h = -5t^2 + 10t -4 has two x-intercepts, namely -.34 and 2.34
Since we only want h ≥ 0, we reject the negative answer.
The parabola doesn't know that it is to represent the height of an object, it
is merely just happy being a parabola.
To find when the ball is 6 meters above the ground, we need to solve the equation -5t^2 + 10t + 2 = 6. This equation represents the height of the ball at any given time t.
First, let's rearrange the equation to make it easier to work with:
-5t^2 + 10t + 2 - 6 = 0
-5t^2 + 10t - 4 = 0
Now, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = -5, b = 10, and c = -4. Plugging these values into the quadratic formula, we get:
t = (-10 ± √(10^2 - 4(-5)(-4))) / (2(-5))
t = (-10 ± √(100 - 80)) / -10
t = (-10 ± √20) / -10
t = -1 ± √(20/10)
t = -1 ± √2
So, there are two possible solutions:
1. t ≈ -1 + √2
2. t ≈ -1 - √2
Now, let's explain why there are two answers. In this context, the ball being 6 meters above the ground can happen twice - once on its way up and once on its way down.
When the ball is thrown up, it reaches a maximum height and then starts to fall back down. At the highest point, the velocity becomes zero. So, the first solution, t ≈ -1 + √2, represents the time when the ball is at 6 meters while going up.
On its way back down, the ball will reach the same height of 6 meters above the ground. This is because the height of the ball only depends on the time and does not consider the direction. Hence, there is a second solution, t ≈ -1 - √2, which represents the time when the ball is at 6 meters while coming down.
To summarize, the two solutions arise because the ball reaches the same height of 6 meters twice - once while going up and once while coming down.