### Block on a Slope

(a) Start with block 1. It starts at rest, then accelerates a distance
of 16 m in a time of 4.2 seconds. Use equation from kinematics:
$x = x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^{2}$
$-16m = (0)+(0)(4.2s)+\frac{1}{2}a_{x}(4.2s)^{2}$
$a_{x} = -1.8\frac{m}{s^{2}}.$
Acceleration of block 2 is just the same as acceleration of block
1! All objects fall with approximately the same acceleration.

(b) We have a relation between final and initial velocities, $v_{x}=-v_{0x}$:
$v_{x} = v_{0x}+a_{x}t$
$-v_{0x} = v_{0x}+a_{x}t$
$-2v_{0x} = (-1.8\frac{m}{s^{2}})(4.2s)$
and so we get $v_{0x}=3.8\frac{m}{s}$.

(c) Half the time is spent coming down from the highest point. So
the time to "fall'' is 2.1s. Distance traveled is again from
kinematic equation:
$x=(0)+(0)(2.1s)+\frac{1}{2}(-1.8\frac{m}{s^{2}})(2.1s)^{2}=-4.0m$.