Answer
(a) $v = \sqrt{2gh}$
(b) The speed at the bottom is 2.4 m/s in both cases.
Work Step by Step
(a) The acceleration down the ramp is $a = g~sin(\theta)$.
We can find the distance $d$ the rock slides;
$d = \frac{h}{sin(\theta)}$
We can find the speed at the bottom;
$v^2 = v_0^2+2ad = 0 +2ad$
$v = \sqrt{2ad} = \sqrt{(2)(g~sin(\theta))(\frac{h}{sin(\theta)})}$
$v = \sqrt{2gh}$
(b) $v = \sqrt{2gh}$
$v = \sqrt{(2)(9.80~m/s^2)(0.30~m)}$
$v = 2.4~m/s$
Since the speed at the bottom does not depend on the angle, the speed at the bottom is 2.4 m/s in both cases.
