derive the equation of motion?

Question

tell me more about the below in relation to the question;
1.maximum height
2..projectile motion

Answer 1

These sorts of issue are all about forces acting on a mass (m). To derive and solve the equations of motion, we must first ID all the forces acting on the mass. To do this, we separate the vertical acting forces (Fv) from the horizontal acting forces (Fh).

In the instance of a cannon ball shot from a cannon, Fh = 0, there are no horizontal forces if we discount drag friction (which we can safely do for a cannon ball). The vertical forces are Fv = mg, which is the force of gravity pulling the ball back towards the Earth. We call this vertical force weight.

Since the only force acting on a ball once it leaves the barrel of the cannon is the ball's own weight, the max height it can reach can be found from PE = mgd = 1/2 mv^2 = KE, where PE and KE are the potential and kinetic energies of the ball. This equation stems from the conservation of energy law. v in this case is the initial vertical velocity, v = V sin(theta), where V is the muzzle velocity and theta is the inclination of the cannon when fired.

Thus, d = 1/2 (v^2)/g which clearly shows that max height depends solely on the square of the initial vertical velocity of the ball. This square factor is important because it says, for example, if I double the muzzle velocity, I can get four times the max height. Such is the nature of kinetic energy.

After the ball reaches d, it starts to fall. Discounting the height of the barrel above ground (which will normally be small compared to d), the impact vertical velocity can be found by v = gt, where t is the time to fall from d to the ground. t can be found from d = 1/2 gt^2; so that t = sqrt(2d/g). (FYI, the v when fired upward will be equal to the v on impact, but in opposite directions. This is why bullets fired into the air are so dangerous when they come back to Earth.)

But, note, this is only the vertical component of the ball's velocity. The ball's horizontal velocity remains constant at h = V cos(theta) because there are no horizontal forces to change it. So the magnitude of the impact is the vector sum of v and h; or I = sqrt(h^2 + v^2). The impact angle (phi) can be found from tan(phi) = v/h.

There are certain simplifying assumptions we typically make in these kinds of physics problem. First, we assume no drag forces, when in fact there may be. Second, we assume the Earth's surface is flat when in fact it is curved. Finally, we discount the so-called Coriolis Effect that can actually bend a trajectory (projectile motion). But for short distances, these assumptions will not significantly change the math results from the actual trajectories.