Answer:

Explanation:
From the question we are told that
Temp of first bolts
Temp of 2nd bolt 
Generally the equation showing the relationship between heat & temperature is given by

Generally heat released by the iron bolt = heat gained by the iron bolt
Generally solving mathematically





Therefore
is the final temperature inside the container
At the time that I'll call ' Q ', the height of the stone that was
dropped from the tower is
H = 50 - (1/2 G Q²) ,
and the height of the stone that was tossed straight up
from the ground is
H = 20Q - (1/2 G Q²) .
The stones meet when them's heights are equal,
so that's the time when
<span>50 - (1/2 G Q²) = 20Q - (1/2 G Q²) .
This is looking like it's going to be easy.
Add </span><span>(1/2 G Q²) to each side.
Then it says
50 = 20Q
Divide each side by 20: 2.5 = Q .
And there we are. The stones pass each other
2.5 seconds
after they are simultaneously launched.
</span>
b). The power depends on the RATE at which work is done.
Power = (Work or Energy) / (time)
So to calculate it, you have to know how much work is done AND how much time that takes.
In part (a), you calculated the amount of work it takes to lift the car from the ground to Point-A. But the question doesn't tell us anywhere how much time that takes. So there's NO WAY to calculate the power needed to do it.
The more power is used, the faster the car is lifted. The less power is used, the slower the car creeps up the first hill. If the people in the car have a lot of time to sit and wait, the car can be dragged from the ground up to Point-A with a very very very small power ... you could do it with a hamster on a treadmill. That would just take a long time, but it could be done if the power is small enough.
Without knowing the time, we can't calculate the power.
...
d). Kinetic energy = (1/2) · (mass) · (speed squared)
On the way up, the car stops when it reaches point-A.
On the way down, the car leaves point-A from "rest".
WHILE it's at point-A, it has <u><em>no speed</em></u>. So it has no (<em>zero</em>) kinetic energy.
Position of B :
x = 4.66*cos 30 = 4.036
y = 3-4.66*sin 30 = 3-2.33 = 0.67
BC = √y^2+(x-1)^2 = √0.67^2+3.036^2 = 3.109
heading = arctan y/(x-1) = arctan 0.67/(3.036) = 12.44° south of west
hope this helps :)