Projectile

Velocity has components $v_y=8-10t, v_x=6$ .

At $t=0$ they are $v_y(0)=8, v_x(0)=6$

Speed is from Pythagorean theorem $\sqrt{8^2+6^2}=\boxed{10}$

The general equation of the height is:
$y=v_{y}\cdot t - \frac{1}{2}gt^{2}$
By compairing this with the equation given we see that:
$v_{y}=8\frac{m}{s}$ and $g=10\frac{m}{s^{2}}$
But: $v_{y}=v\cdot sin{\theta}$
Therefore: $v\cdot \sin{\theta}=8\Rightarrow v^{2}(\sin{\theta})^{2}=64$ ...... (1)

The general equation for the horizontal distance is:
$x=v_{x}\cdot t$
By compairing this with the equation given we see that:
$v_{x}=6\frac{m}{s}$
But: $v_{x}=v\cdot \cos{\theta}$
Therefore: $v\cdot \cos{\theta}=6\Rightarrow v^{2}(\cos{\theta})^{2}=36$ ...... (2)
By adding (1) and (2) we get:
$v^{2}(\sin^{2}{\theta}+\cos^{2}{\theta})=100\Rightarrow v^{2}=100\Rightarrow v=10\frac{m}{s}$

Notation : $v$ is the initial speed, $v_{x}$ is the inital speed on the x-axis, $v_{y}$ is the inital speed on the y-axis, $t$ is time, $g$ is the acceleration of gravity, $\theta$ is the initial angle at which we throw the object