a rubber ball is thrown upwards with an initial speed of 20m/s. The approximate height o the ball of t seconds is h(t)= 20t-4.9t^{2}. Determine the time period, to the nearest hundredth of a second, during which the ball is higher than 8 m.

6 years ago

Answered By Vincent E

What you have there is an expression for the height of the ball. $h\left(t\right)=-4.9t^2+20t$h(t)=−4.9t^{2}+20t .

We want to find the time period where the height of the ball is higher than 8m.

If we subsitute 8m into our expression for the height, we obtain a quadratic equation. We can then find the roots of this quadratic equation using the quadratic formula $t=\frac{\left(-b\pm\sqrt{b^2-4ac}\right)}{2a}$t=(−b±√b^{2}−4ac)2a or by using a graphing calculator and inspecting the graph. You will find two roots. Between those two roots (or times in this case), the ball will be higher than 8m.

If you want to use a graphing calculator, there's two ways. You can either graph the $h\left(t\right)=-4.9t^2+20t$h(t)=−4.9t^{2}+20t in Y1 and h=8 in Y2 and then search for the intersect. Or you can just graph $-4.9t^2+20t-8$−4.9t^{2}+20t−8 and calculate the roots. Both will give you the same result.

6 years ago

## Answered By Vincent E

What you have there is an expression for the height of the ball. $h\left(t\right)=-4.9t^2+20t$h(t)=−4.9t

^{2}+20t .We want to find the time period where the height of the ball is higher than 8m.

If we subsitute 8m into our expression for the height, we obtain a quadratic equation. We can then find the roots of this quadratic equation using the quadratic formula $t=\frac{\left(-b\pm\sqrt{b^2-4ac}\right)}{2a}$t=(−b±√b

^{2}−4ac)2a or by using a graphing calculator and inspecting the graph. You will find two roots. Between those two roots (or times in this case), the ball will be higher than 8m.If you want to use a graphing calculator, there's two ways. You can either graph the $h\left(t\right)=-4.9t^2+20t$h(t)=−4.9t

^{2}+20t in Y1 and h=8 in Y2 and then search for the intersect. Or you can just graph $-4.9t^2+20t-8$−4.9t^{2}+20t−8 and calculate the roots. Both will give you the same result.