For a 2D projectile, how do I calculate the maximum height of the projectile, if gravity is the only acceleration acting on the projectile?
First, the time it takes for the projectile to hit the ground needs to be found. Assuming the initial velocity vector is known, this time may be calculated by solving the equation yfinal=y0+Vy0*T-0.5*g*t^2 for variable t. Since there is no horizontal acceleration, the horizontal velocity of the projectile will be equivalent to the initial horizontal velocity; that is to say, the horizontal velocity is constant. Therefore, the projectile will reach its maximum height at exactly t/2, where t is the value of time that it took for the projectile to hit the ground. Therefore, t/2 may be plugged back into y(t) to get the maximum height of the projectile.
I do not understand the 2nd law of thermodynamics.
The second law of thermodynamics is directly related to the entropy of a closed system. Essentially, it says that the quality of energy degrades as it changes from one form to another. This degradation is irreversible.
Given a function f(x), I do not understand how to find its maximum and minimum values.
When f(x) is at a maximum or minimum, its slope will be zero. Knowing that the derivative of f(x), which may be denoted f'(x), yields the value of the slope of f(x) for any x, we can take the derivative of f(x) and solve for the value(s) of x that make f'(x)=0. Each of these x values that satisfy f'(x)=0 may then be plugged back into f(x) to get the maximum and minimum values of f(x).