For Item #17:
There are two methods to approach this type of problem.
If you like using sin, cos, and tan, use method 1:
Sketch the right triangle where the height is 20 m and the hypotenuse is 40 m. The angle above the horizontal (theta) is therefore 30 degrees (sin theta = 20/40). The rate of acceleration down a frictionless ramp of an angle theta above the horizontal is: g sin theta (where g is rate of acceleration due to gravity in free fall). Thus, the rate of acceleration down the ramp is 10 sin 30 = 5 m/s^2.
Now, use the kinematic equation:
d = (vi)(t) + (1/2)(a)(t^2)
Where d is distance (40 m), vi is initial velocity (0 m/s), t is time, a is rate of acceleration (5 m/s^2):
40 = (0)(t) + (1/2)(5)(t^2)
40 = (2.5)(t^2)
16 = t^2
t = 4 seconds. Therefore, the correct answer is C (4 seconds).
If you don't like using sin, cos, and tan, use method 2:
Calculate how long it would take to free-fall from its height of 20 m by using the kinematic equation:
d = (vi)(t) + (1/2)(a)(t^2)
Where d is distance (-20 m), vi is initial velocity (0 m/s), t is time, a is rate of acceleration (-10 m/s^2):
-20 = (0)(t) + (1/2)(-10)(t^2)
-20 = -(5)(t^2)
20 = (5)(t^2)
4 = t^2
t = 2 seconds
Next, set up a ratio:
distance of ramp/distance of free-fall = time down ramp/time of free-fall
40/20 = t/2
t = 4 seconds (option C)
This works due to similar triangles and ratios (it is where "g sin theta" from method 1 is derived as well).
I hope one of those methods makes sense for you.
Good Luck!