Answer
The speed at the bottom of the incline is $6.0~m/s$
Work Step by Step
We can use conservation of energy to find the speed at the bottom of the incline:
$KE_f +U_f= U_0+KE_0$
$\frac{1}{2}mv_f^2+0 = mgh+\frac{1}{2}mv_0^2$
$v_f^2 = 2gh+v_0^2$
$v_f = \sqrt{2gh+v_0^2}$
$v_f = \sqrt{(2)(9.80~m/s^2)(2.0~m)~sin~30.0^{\circ}+(4.0~m/s)^2}$
$v_f = 6.0~m/s$
The speed at the bottom of the incline is $6.0~m/s$.
