Given that the acceleration due to gravity is 32.2 feet per second per second. Find theheight above the ground of an object thrown upward from a pointS0meters abovethe ground with an initial velocity ofv0meters per second. Hint: Acceleration due togravity is in the downward direction, and therefore will be negative

We have that:

The acceleration is a = 32.2 feet per second or 9.8 m/s^2 (we should use this one because the other numbers are in meters)

S0 is the initial position

V0 is the initial velocity.

We start for the acceleration, we have:

A(t)  -9.8m/s^2

for the velocity, we integrate over time:

V(t) = ( -9.8m/s^2)*t + V0

for the position we integrate again:

P(t) = (1/2)( -9.8m/s^2)*t^2 + V0*t + S0

P(t) is the height above the ground as a function of t, we can find the maximun height by findind at which time the velocity is equal to zero (this means that the object stops going up and starts falling down)

v(t) = 0 = ( -9.8m/s^2)*t + V0

t = V0/( -9.8m/s^2)

Now, we replace this in the position equation and we can find the maximum height,

maxP = (1/2)( -9.8m/s^2)*(V0/( -9.8m/s^2))^2 + V0^2/( -9.8m/s^2) + S0

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

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