Calculus

Given that dv/dt = a, you can take the integral of a, and get...

vt = a t + C. Substitute v0 for C, and you get vt = a t + v0

Assuming that the acceleration due to gravity is about -32 fps2, then the velocity after 2 seconds of an object down with an initial velocity of -20 fps is -80 fps.

To solve the second part of the question, "what is the velocity after falling 150 f?", you need to integrate again, and get...

xt = 1/2 a t2 + v0 t + C. Substitute x0 for C, and you get xt = 1/2 a t2 + v0 t + x0

Solve for t, with xt = 150, (300-150), v0 = -20, and x0 = 300, using the quadratic equation, and you get t = 2.28 s. (You also get t = -1.03 s, but you can ignore that because it is the solution for time prior to the release point.)

Go back to the first integral and plug in t = 2.28 s, and you get v = 92.9 fps.

This answer assumes no resistance from velocity in air. At 92.9 fps, you might get some effect, depending on the shape and density of the object, so the real answer is less than stated.

1 + cos(x) = sin(x)

First of all, since you didn't include the argument of the sin or cos in the question,

we sadly suspect that you have little clue as to what you're looking for, or what a

solution will look like.

==> You need to find an angle whose sine is 1 greater than its cosine.

The numerical values of both the sine and cosine functions range from -1 to +1.

No angle has a sin or cosine less than -1 or greater than +1. That'll help us put

some constraints on the equation, and see what may be going on.

The equation also says: sin(x) - cos(x) = 1

This would be a great place to flash a sketch of the graphs of the sin(x) and cos(x)

functions up on the screen, and see where they differ by roughly 1, with the sine

being the greater one. It's too bad that we can't do that. The best we can do is to

draw them on our scratch pad here, look at them, and tell you what we see:

-- The sine is greater than the cosine only between 45° and 225°,

so any solutions must be in that range of angles.

-- At 90°, the sine is 1 and the cosine is zero, so we have [ 1 + 0 = 1 ], and 90° definitely works.

-- At 180°, the sine is zero and the cosine is -1, we have [ 1 + -1 = 0 ], and 180° works.

-- If there were any range between 45° and 225° where the graphs of the sine

and cosine functions were parallel curves, then any angle in that range might

also be a solution.

But there isn't any such place. 90° and 180° are the only points where the values

are different by 1 and the sine is greater, so those are the only principle solutions

(answers between zero and 360°.)