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.