A 200 kg boulder is 1000m above the ground. What is the speed of the object right before it hits the ground?

Explanation:

we use the equation for the final speed:

where is the final speed, is the initial speed, is the gravitational acceleration () and is the height.

In this case we don't have an initial velocity indicated so: , and we are told that the boulder is at a height: .

We substitute this values into the equation:

the speed of the object right before it hits the ground is

c.the charges if the proton and electron and gravity

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west