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.
set h=6
rearrange the quadratic equation to get
5t^2 - 10t + 4=0
Solve using the formula, you should get two positive answers, one will represent the time for its upwards path, and the second the time on its downwards trip.
To find when the ball is 6m above the ground, we need to set the height equation h equal to 6:
-5t^2 + 10t + 2 = 6
Next, rearrange the equation to get it in standard quadratic form:
-5t^2 + 10t - 4 = 0
Now, we can solve this quadratic equation to find the values of t. We can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values from our equation, we get:
t = (-10 ± √(10^2 - 4(-5)(-4))) / (2(-5))
Simplifying further:
t = (-10 ± √(100 - 80)) / (-10)
t = (-10 ± √20) / (-10)
t = (-10 ± 2√5) / (-10)
This gives us two possible solutions:
t = (-10 + 2√5)/(-10) and t = (-10 - 2√5)/(-10)
Now, let's simplify those values:
t ≈ 1.5274 and t ≈ 0.47256
Since time cannot be negative, we ignore the negative solutions. Therefore, the two positive solutions are approximately 1.53 seconds and 0.47 seconds.
These two solutions represent the time when the ball is 6m above the ground. The first solution, 1.53 seconds, represents the time during the upward path of the ball when it reaches a height of 6m. The second solution, 0.47 seconds, represents the time during the downward path of the ball when it reaches a height of 6m.
This is why there are two answers to the question of when the ball is 6m above the ground.