Why the horizontal component in a projectile always remaIns constant?

Answer 1

the horizontal velocity of a projectile remains constantbecause, there is no horizontal force acting on it, so no acceleration in the horizontal direction.

Answer 2

Newtons 1st Law: When no net force acts on a body its velocity can not change.

There are horizontal forces acting on a projectile but it is far from trivial to include them and since these forces are sometimes quite small by design they can be ignored until later in most mechanics and physics courses. This allows the student to become comfortable with components of vectors and the associated calculations.

PeterVincent

Answer 3

The vertical componant is being acted upon by the force of gravity, pulling it downward toward the Earth.

There is no external force (assuming no wind or drag) acting on a projectile in the horizontal direction.

Answer 4

if you aren't from now on confident about those calculations, watch the video-training contained in the hyperlinks - they conceal projectiles from the basics. that's numerous labor to respond to the completed question, with causes, yet the following is going... _______________________________________ A) The horizontal component of speed is consistent (as there isn't any horizontal rigidity) and is 40m/s. (The '4 seconds' is a pink herring.) If the launch p.c. = V, its horizontal ingredient is Vcos(30?) Vcos(30?) = 40 V = 40/0.866 = 40 six.2m/s = 46m/s to 2 major figures _______________________________________... B) The vertical component of speed is Vsin(30?) = 40 six.2 x 0. 5 = 23.1m/s pondering vertical action as a lot as max authentic, and taking upwards as useful: preliminary vertical component of speed, u = 23.1m/s very last vertical component of speed, v = 0 (at max authentic) acceleration = g = -9.8m/s² v = u+at 0 = 23.a million +(-9.8) t t = 23.a million/9.8 = 2.36s = 2.4s to 2 major figures. _______________________________________... C) i guess you propose t=4s. The projectile should be falling because it reached max authentic beforehand (at t=2.4s) as calculated partly B. _______________________________________... D) It took 2.36s to achieve max authentic. because the path is symmetrical, that's going to take yet another 2.36s to fall again. finished flight time = 2 x 2.36 = 4.72s With a consistent horizontal p.c. of 40m/s, variety = p.c. x time = 40 x 4.seventy 2 = 189m (190m to 2 major figures). _______________________________________... E) no longer confident what they want the following. the most direct answer is basically to workout recurring the recent variety. preliminary horizontal component of speed = 40 six.2cos(6?) = 35.9m/s preliminary vertical component of speed = 40 six.2sin(6?) = 4.83m/s Time to achieve max authentic (using an similar approach as section B) is given by: 0 = 4.eighty 3 + (-9.8)t t = 0.49s finished flight time = 2 x 0.49s = 0.98s variety (using similar approach as section D) = 35.9 x 0.ninety 8 = 35m So the range has decreased. Herre is yet another approach: optimum variety takes position even as the perspective of projection is 40 5?. increasing the perspective above 40 5? decreases the range. lowering the perspective below 40 5? dereases the range. 6? is a larger decrement from 40 5? than 30? is, so the range will be below even as the perspective is 30?.

Answer 5

In reality it doesn't, the horizontal velocity gets less as it flies, but that makes the simple physics problems so complicated that the assumption is made that it is constant which is pretty much true unless you are firing artillery shells 18 miles and 20,000 feet high. Air drag varies with velocity, with altitude (air density), humidity, etc.

Answer 6

Once the projectile leaves the gun or force pushing it there is no longer anything accelerating it.

Really they slow down due to air resistance but the maths is too difficult for most undergraduates so air resistance is ignored.

Answer 7

EASY, because there is no acceleration on the horizontal component.