Answer:
The quantity of electrons that flows past a given point is 3.0 C.
Explanation:
An electric current (I) is the ratio of the quantity of charges (Q) that flows through a point to the time taken (t).
i.e I = 
It is measured in Ampere's by the use of an ammeter in the laboratory. The quantity of charge that flow through a given point is measured in Coulombs, while time is measured in seconds.
Given that; I = 1.5A and t = 2s, find Q.
Q = It
= 1.5 × 2
= 3.0 C
The quantity of electrons that flows past a given point is 3.0 C.
The answer is approximately 2.998e+8
Ok, this is a 2d kinematics problem, the falls 14 m part is confusing, I think it means in the x direction, but you don't need it anyway.
If we know it goes 4m into the air, we know d = 4m (height of wall), we also know the acceleration a=-9.8m/s^2 (because gravity) and that the vertical velocity when it just clears the wall will be 0 m/s, which we'll call our final velocity (Vf). Using Vf^2 = Vi^2 +2a*d, we can solve this for Vi and drop Vf because it's zero to get: Vi = sqrt(-2ad), plug in numbers (don't forget a is negative) and you get 8.85 m/s in the vertical direction. The x-direction velocity requires that we solve the y-direction for time, using Vf= Vi + at, we solve for t, getting t= -Vi/a, plug in numbers t= -8.85/-9.8 = 0.9 s. Now we can use the simple v = d/t (because x-direction has no acceleration (a=0)), and plug in the distance to the wall and the time it takes to get there v = (4/.9) = 4.444 m/s, this is the velocity in the x direction, we use Pythagoras' theorem to find the total velocity, Vtotal = sqrt(Vx^2 + Vy^2), so Vtotal = sqrt(8.85^2+4.444^2) = 9.9m/s. Yay physics!
