a ball is dropped from rest at a height of 89m above the ground. (a)what is it's speed just before it hits the ground? (b) how long does it take to reach the ground?​

(a) 41.75m/s

(b) 4.26s

Explanation:

Let:

 Distance, D = 89m

 Gravity, = 9.8 m/

Initial Velocity, = 0m/s

Final Velocity, = ?

Time Taken, = ?

With the distance formula, which is

D = +

and by substituting what we already know, we have:

89 = ×9.8×

With the equation above, we can solve for :

Now that we have solved , we can use the following velocity formula to solve for :

, where is also equals to , so we have

By substituting , , and ,

We have:

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).