2. Projectile Motion - Mechanics Kinematics - The Scientific Sentence

Kinematics: projectile in two dimensions motion

1. Definitions:

A projectile is defined as an object thrown from a certain surface (Earth) moving against a gravity. A golf ball is a projectile. The effect (how it behaves) of the motion of a projectile is caused, set and dictated by the presence of gravity. The gravitational acceleration "g" (deceleration) is a vertical acceleration, and directed downward. The projectile motion is not decelerated because there is a change in the velocity, it is accelerated because it is present in the gravity. That implies the change in the velocity Δ⃗v. ( "g" exists → Δ⃗v exists). Since ⃗g is directed vertically downward, so does Δ⃗v. Thus there is no horizontal component for the acceleration. If ⃗a is the acceleration of the object, ⃗a = ⃗ax + ⃗ay = 0 + ⃗ay = ⃗ay = ay ⃗j = - ⃗g = - g ⃗j = - 9.81 ⃗j.

2. Projectile motion: ballistic trajectory

If ⃗ax = 0, that is d⃗vx/dt = 0, then: ⃗vx = constant = ⃗vxo. Hence: ⃗x = ⃗vxo t + ⃗xo.At t = 0, we have ⃗x = o, thus ⃗xo = 0, Hence:The component of the projectile motion, along the "x" axis is:⃗x = ⃗vxo t = vxot ⃗iSimilarly, If ⃗ay = - ⃗g = constant, that is d⃗vy/dt = constant, then: ⃗vy = - ⃗g t + constant = - ⃗g t + ⃗vyo. Hence: ⃗y = - (1/2) ⃗g t2 + (⃗vyo) t + constant2 = = - (1/2) ⃗g t2 + (⃗vyo) t + ⃗yo.At t = 0, we have ⃗y = o, thus ⃗yo = 0, Hence:The component of the projectile motion, along the "y" axis is:⃗y = - (1/2) ⃗gt2 + (⃗vyo) t = - (1/2) g t2 ⃗j + (vyo t) ⃗jEliminating the parameter time "t" in the two above equations can be as follows: From the first equation, we have t = x/ vxo. Substituted in the second equation, gives:y = - (1/2) g (x/ vxo)2 + vyo (x/ vxo) = - [(1/2) g / vxo2] x2 + [vyo/ vxo] x.y = - [(1/2) g / vxo2] x2 + [vyo/ vxo] xWhich is an equation of a parabola.The range of the path (parabola) corresponds to: - [(1/2) g / vxo2] x2 + [vyo/ vxo] x = 0; that gives: xr = 2[vyo vxo]/gThe range is: xr = 2vyo vxo/gThe maximum corresponds to:dy/dx = 0, that is - [ g / vxo2] x + [vyo/ vxo] = 0. Hence:x = vyo vxo / g , and y = y = - [(1/2) g / vxo2] (vyo vxo / g)2 + [vyo/ vxo] (vyo vxo / g) =- (1/2) (vyo 2/g) + vyo2/g = (1/2) vyo 2 / gCoordinates of the maximum point (height of the trajectory):xm = vyo vxo/g         ym = vyo 2/2g

3. Projectile motion: polar coordinates

In polar coordinates, we have:⃗x = ⃗vxot = (vo cosθ t) ⃗i⃗y = - (1/2) g t2 ⃗j + (vo sinθ t) ⃗jy = - [(1/2) g / vo2 cos2θ] x2 + [vo tgθ] xxm = vo2 cosθ sinθ / g = (vo2/2g) sin 2θym = (vo 2/2g) sin2 θAt θ = 45o:⃗x = (21/2/2)vo t ⃗i⃗y = - (g/2) t2 ⃗j + (21/2/2)vo t ⃗jy = - [(g/4) / vo2] x2 + (1/2)vo xxm = vo2/2gym = vo 2/4g

4. Projectile Motion: Particular cases:

4.1. One dimension motion: elevation θ = 0

For a horizontal launch: ⃗x = ⃗vxot = vot ⃗i. The velocity ⃗v = d⃗x/dt = vo⃗i is constant; and the acceleration d⃗v/dt is null.

4.2. One dimension motion: elevation θ = π/2

For a vertical launch:⃗x = 0⃗y = [- (1/2) g t2 + vot] ⃗jxm = 0ym = vo 2/2gy = - (1/2) g t2 + votym = vo 2/2gThe time of flight tf for this vertical launch can be found as follows:y = ym = h = - (1/2) g t2 + votThis quadratic equation can be written:(1/2) g t2 - vot + h = 0Which has the following two roots:tf1 = [vo + (vo2 - 2gh)1/2]/2tf2 = [vo - (vo2 - 2gh)1/2]/2Limit case:(vo2 - 2gh)1/2 = 0. That is vo = (2gh)1/2Then: t = vo/g, that corresponds to the height of the trajectory.

4.3. Two dimensions motion: Free-fall with initial velocity:

At the height "h", if an an object is launched with a velocity ⃗vo directed at an angle θ with respect to the horizontal line, this object undergoes a free-fall effect.If we decompose the vector velocity ⃗vo into two perpendicular components ⃗voxt = ⃗vo cos θ and ⃗voy = ⃗vo sin θ, we can write:- Along the "x" axis:x(0) = 0, the initial velocity is ⃗v0x = v0x ⃗i = v0cos θ ⃗i, and there is no acceleration ⃗ax = 0. Hence:⃗x(t) = ⃗voxtx(t) = voxt = vo cos θ t- Along the "y" axis:y(0) = h, the initial velocity is ⃗v0y = v0y ⃗j = v0sin θ ⃗j, and the acceleration is imposed by gravity: ⃗ay = - g⃗j. Hence:⃗y(t) = [h + voy t - (1/2) g t2]⃗jy(t) = h + voy t - (1/2) g t2 = h + vo sinθ t - (1/2) g t2y(t) = h + vo sinθ t - (1/2) g t2The period of time lasted by the motion can be found as follows:y(0) = 0, that is:h + vo sinθ tl - (1/2) g tl2] = 0(1/2) g tl2 - vo sinθ tl - h = 0This quadratic equation has the following root:tl = ( vo + [vo2 sin2θ + 2gh]1/2)/g;The range is then equal to: xr = tl vo cosθ

4.4. Two dimensions motion: Range and maximum height:

1. Along the x-axis:acceleration a = 0xo = 0x = vox t = vo cosθ tx = vox t = vo cosθ t2. Along the y-axis:acceleration a = - g = - 9.80 m/s2yo = hoy = yo + voy - (1/2)g t2 = ho + vosin θ - (1/2)g t2y = ho + vosin θ t - (1/2)g t2y = ho + vosin θ t - (1/2)g t23. At hmax:we have vx = vox, and vy = 0, so02 = voy2 + 2 (- g) (hmax - ho), that is:0 = voy2 - 2 g (hmax - ho). Therefore:hmax = [(vosin θ)2/2g] + hohmax = [(vosin θ)2/2g] + hohmax = [vosin θ)2/2g] + ho4. At x = r:y = 0 , then:y = ho + vosin θt - (1/2)g t2 = 04.1. Simple method:Solve for the time t. We find t+ and t- .The convenient solution is t = t+.Using t+, we find:x = r = vox t+ = vocosθ t+The range is r = vocosθ t+r = vocosθ t+4.2. Other method:substitute x = voxt = vo cos θ t in the equation:y = yo + vo sinθ t - (1/2) g t2 , and find:y = yo + tanθ x - (1/2) g (1/vocos θ)2 x2When y = 0, we solve for x the quadratic following equation:0 = yo + tanθ x - (1/2) g (1/vocos θ)2 x2, and find the range x = r.In the case of yo = 0 , we have:0 = tanθ x - (1/2) g (1/vocos θ)2 x2Hence:x = 0, ortan θ - (1/2) g (1/vocos θ)2 xThat is:x = r = 2 tan θ (vocos θ)2/gx = r = 2 tan θ (vocos θ)2/g

5. Particular cases:

5.1. The initial velocity is null:

vo = 0 we have a free-fall motion.The equation of the motion is:y(t) = h - (1/2) g t2 = h + vo sinθ t - (1/2) g t2.The period of time lasted by the motion is determined as follows:y(t) = h - (1/2) g t2 = 0.Then:tl = (2h/g)1/2The range is then equal to: xr = 0

5.2. The initial velocity is horizontal:

voy = 0 , then vox = vo. We have also a free-fall motion.The equation of the motion is:y(t) = h - (1/2) g t2Equating to zero: y(t) = h - (1/2) g t2 = 0 gives the period of time lasted by the motion:tl = (2h/g)1/2The range is then equal to: xr = tlvo = (2h/g)1/2vo

5.3. The initial velocity is vertical:

vox = 0 , then voy = vo. We have also a free-fall motion.The equation of the motion is:y(t) = h + vot - (1/2) g t2Equating to zero: y(t) = h + vo - (1/2) g t2, that is (1/2) g t2 - vo - h = 0 gives the period of time lasted by the motion:tl = ( vo + [vo2 + 2gh]1/2)/gThe range is then equal to: xr = tlvoRemark:The period of time lasted by the motion is the same whatever the initial velocity is null or horizontal.

5.4. One dimension motion: Mixed: Free-fall and vertical initial velocity:

The object A moves upward with a velocity VAo. The Kinematic equations give for its position :yA (t) = 0 + VAo t - (1/2) g t2. For the object B, we have : yB (t) = h + 0 - (1/2) g t2.At a certain time tm, the two objects will have the same position from the ground (axis x). We can then write:yA (tm) = yB (tm); that gives:VAo tm - (1/2) g tm2 = h + 0 - (1/2) g tm2, that yield:VAo tm = h , then:tm = h /VAoThe same position has the value:yA = yB = h - (1/2) g [h /VAo]2It could happen that the object A turns back before meeting the object B. A turns back at the time given by zeroing the derivative of the expression of its position: yA (t) = VAo t - (1/2) g t2; that is:d[yA (t)]/dt = VAo - g tb = 0 , that yields:tb = VAo / g In fact, they will meet at least if tb is equal or greater than tm that is : tb >= tm, then:VAo / g >= h /VAo, or VAo2 >= gh . That is:VAo >= [gh ]1/2Example:If h = 80 m, we have: VAo >= [9.81 x 80 ]1/2 = 28 m/s.An initial speed less than this value is not sufficient for the object A to meet the object B. The latter will turn back without meeting the former and fall on the ground before it does.

projectile gravity free-fall range deceleration