Answer
The puck needs a minimum speed of 4.5 m/s
Work Step by Step
Let $d$ be the length of the ramp. We can find the height $h$ of the ramp.
$\frac{h}{d} = sin(\theta)$
$h = d~sin(\theta)$
$h = (3.0~m)~sin(20^{\circ})$
$h = 1.03~m$
The puck's minimum kinetic energy at the bottom of the ramp should be equal to the potential energy at the top of the ramp.
$KE = PE$
$\frac{1}{2}mv^2 = mgh$
$v^2 = 2gh$
$v = \sqrt{2gh}$
$v = \sqrt{(2)(9.80~m/s^2)(1.03~m)}$
$v = 4.5~m/s$
The puck needs a minimum speed of 4.5 m/s.
