Answer
At $~~t=2.2~s~~$ the speed is $10~m/s$
Work Step by Step
$v_y$ is constant and equal to $8.0~m/s$
We can find the required value of $v_x$:
$v = \sqrt{v_x^2+v_y^2} = 10~m/s$
$v_x^2+v_y^2 = (10~m/s)^2$
$v_x^2 = (10~m/s)^2-v_y^2$
$v_x = \sqrt{(10~m/s)^2-v_y^2}$
$v_x = \sqrt{(10~m/s)^2-(8.0~m/s)^2}$
$v_x = \pm 6.0~m/s$
We can try to find $t$ when $v_x = 6.0~m/s$:
$6.0t-4.0t^2 = 6.0$
$4.0t^2-6.0t+ 6.0 = 0$
$2.0t^2-3.0t+ 3.0 = 0$
We can use the quadratic formula:
$t = \frac{3.0 \pm \sqrt{(-3.0)^2-(4)(2.0)(3.0)}}{(2)(2.0)}$
$t = \frac{3.0 \pm \sqrt{-15}}{4.0}$
There are no solutions.
We can find $t$ when $v_x = -6.0~m/s$:
$6.0t-4.0t^2 = -6.0$
$4.0t^2-6.0t- 6.0 = 0$
$2.0t^2-3.0t- 3.0 = 0$
We can use the quadratic formula:
$t = \frac{3.0 \pm \sqrt{(-3.0)^2-(4)(2.0)(-3.0)}}{(2)(2.0)}$
$t = \frac{3.0 \pm \sqrt{33}}{4.0}$
$t = -0.69~s, 2.2s$
At $~~t=2.2~s~~$ the speed is $10~m/s$
