how to study projectile motion? how to decide which formula must be applied?

Answer 1

I assume you mean projectile motion on earth, and that the object will go through the air. Examples are (a) dropping a ball from the top of a building, (b) throwing a ball straight up in the air, (c) throwing a ball horizontally or at an angle from the top of a cliff, (d) throwing a ball as far as you can on a level field, or (e) throwing a ball to strike the side of a building.

In all five cases, gravity works on the ball. In examples (c) through (e), there's also a horizontal component, but not in examples (a) or (b).

Once the ball is in motion, gravity is the only force operating, and from Newton's law F = ma, we get the constant acceleration due to gravity, a = F/m = g, where g = - 32 ft/sec/sec in the English system, or - 9.8 m/sec/sec in SI units. (g is the constant acceleration due to gravity.)

From that, velocity v = gt + vo, where t is time and vo is the initial velocity.

From that, displacement s = 1/2 gt^2 + vo t + so, where so is initial displacement.

If the motion is vertical only, the above formulas are all you need. Dropping a ball from the top of a building (case a), there is some initial displacement (the height of the building), but the initial velocity is zero.

If you throw a ball straight up (case b), there is an initial velocity (feet per second, or whatever) but no initial displacement.

When there's a horizontal component (cases c, d, e), then you need to split the problem in two parts -- a horizontal component and a vertical component. You have to split up the "initial conditions", particularly the initial velocity. Sometimes you have to use trig (sines and cosines) to do this. It depends on the angles you use.

Once you have the problem split into horizontal and vertical components, just use the above formulas for the vertical part subject to gravity. For the horizontal part, the horizontal velocity will be constant, that is, vx = vox (initial horizontal velocity along x-axis); and horizontal displacement sx = vox t + sox where sox is initial horizontal displacement along the x-axis.

Here's a fairly comprehensive example from a physics book:

A stone is thrown from the top of a building upward at an angle of 30 degrees to the horizontal and with an initial speed of 20 m/s. If the height of the building is 45 m, (a) how long is the stone "in flight"? (b) What is the speed of the stone just before it hits the ground? (c) Where does the stone strike the ground?

Start by getting the horizontal and vertical components of the initial velocity. Since it's at a 30 degree angle, vox = 20 cos 30 = 20 [sqrt(3)/2] = 17.3 m/s. (You get this cosine from the 30-60-90 special triangle.) voy = 20 sin 30 = 20(1/2) = 10 m/s.

Next, how long does it stay in the air? Use the vertical component only. Initial vertical displacement soy is +45, and the formula is

s = 0 = (1/2)(-9.8 m/s/s) t^2 + 10 m/s t + 45 m

where the final displacement (the ground) is s=0; g = -9.8; and voy = 10 m/s.

This gives a quadratic

4.9 t^2 - 10 t - 45 = 0

Solve it using the qauadratic formula to get a positive root at t = 4.22 seconds. That's how long the stone stays in the air before hitting the ground. (Answer a.)

To get the speed of the stone, use the vertical velocity formula
vy = -gt + voy = (-9.8 m/s/s)(4.22 s) + 10 m/s = -31.4 m/s

(The negative sign indicates the stone is falling downward.)

There's a constant horizontal component to the velocity also. vx = vox = 17.3 m/s. Since horizontal and vertical components are at right angles to each other, use the Pythagorean Theorem to get the final speed of the stone:

v = sqrt[(-31.4)^2 + 17.3^2] = 35.9 m/s (answer b)

That's how fast the stone is going when it hits the ground.

The last question is, how far out did it go? That's the horizontal displacement sx. Since horizontal velocity is constant, we only have to multiply by the time in flight:

sx = (17.3 m/s)(4.22 s) = 73 m (answer c)

The stone landed 73 m from the base of the building.

Hope this tells you everything about projectile motion.

Answer 2

The projectile action is rather elementary for air and water. All you pick is the Reynolds variety (density*vel*function length/viscosity) and the dynamic stress (0.5 * Density *velocty). The Reynolds variety and shape (and mach variety if going quickly sufficient) provides you with a coefficient(could be stumbled on for many shapes or variety of approximated by using countless techniques). Muliply that by using the dynamic stress and the section and you gets all the forces genuine rapid. upload in gravity ,then do basically numerical integration and you will create a trajectory. The interface of water to air is a tricky section. the different difficult section is that if the article is shifting too quickly interior the water it is going to cavitate( create air bubbles), and throw all the above steps out the window. a sturdy e book on the basics could be something like fundamentals of Aerodynamics by using Anderson. it somewhat is properly written and rather elementary to study. McCormick wrote a sturdy one stated as aerodynamics to boot. a sturdy e book on numerical techniques might help in case you pick to do it properly too.

Answer 3

You will find the answer here sahil...take care x