a ball is thrown straight up its height h(metres) after t seconds is given, h=-5t^2+10t+2. To the nearest tenth of a second, when is the ball 6m above the ground? Explain why there is two answers. Help me please
Hello, mcinnes!
It's simple algebra . . . Exactly where is your difficulty?
The question is: When is $\displaystyle h = 6$ ?A ball is thrown straight up.
Its height $\displaystyle h$ in metres after $\displaystyle t$ seconds is: .$\displaystyle h \:=\:-5t^2+10t+2$
To the nearest tenth of a second, when is the ball 6m above the ground?
Explain why there is two answers.
We have: .$\displaystyle -5t^2 + 10t + 2 \:=\:6 \quad\Rightarrow\quad 5t^2 - 10t +4 \:=\:0$
Quadratic Formula: .$\displaystyle t \;=\;\frac{10 \pm\sqrt{20}}{10} \;=\;\frac{5\pm\sqrt{5}}{5} \;\approx\;\begin{Bmatrix}1.4 \\ 0.6\end{Bmatrix}$ seconds.
It attains a height of 6 meters on the way up and on the way down.
im doing math on my own, and need to see how some questions are answered, with all the work, and what not, so i can do them on all the other questions i have, with this one, i dont get the two answer part of it, i wouldnt mind knowing why or how there is two answers:P