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# Two Dimensions Motion Assignment Help

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Classical Physics - Two Dimensions Motion

**Two Dimensions Motion**

The best way to solve a two dimensional problem is to convert it into two. One dimensional motion, Say along x- and y- directions. Thus resolve the velocity v to its x and y components say

**V = v**_{xi }+ v_{yj}.

Treat the motion in x- and y- directions as one dimensional motion. The acceleration can also be resolved and taken separately along **x- and y- **directions.

Projectile a freely falling body having constant horizontal velocity may be termed as a projectile. In general, we can say that if velocity along one-direction, that is, x-direction remains constant and there exists along**y-**direction an accelerated motion, then a body undergoing such a motion is called a projectile. It always follows a parabolic path consider a projectile fired with a velocity v making an angle θ with the horizontal from the point 0 (considered origin) at** t = 0**. resolving **v** along** x – **and **y-**directions

vx = v cos θ

And vy – v sin θ

Maximum height attained (hmax) can be calculated by assuming that at maximum height, the vertical component of the velocity become zero.

That is,

Applying **v**^{2} – u^{2} = 2as, we get v^{2}y – v^{2}y = - 28h_{max}

Or **h**_{max} = v^{2} sin^{2} θ / 2g as v’y = 0

Time of flight** (T) **we calculate the time in which the projectile reaches the highest point, when we double this time it would be the time of flight as it would take same time (whatever it takes to each the highest point) to come back on the ground level at** Q**.

Applying **v = u + at**

Let t be the time to reach highest point

**v’y = vy – gt or t = v sin θ/8**

Time of flight **T = 2t = 2v sin θ/8**

Thus the time of flight is the time taken by the projectile to return to the ground from the instant of projection or it is the time for which the projectile remains in air above the horizontal plane from the point of projection.

Horizontal range** (R)** the distance covered along the horizontal direction between the point of projection and point of return on the ground is called horizontal range.

**R = vx x time of flight = v cos θ (2v sin θ / 8) = v**^{2} sin^{2}θ/8

**R**_{max }(maximum range) Rmax will be obtained when **sin**^{2}θ = 1

That is, **2θ = 900 or θ = 450 and Rmax = v**^{2} / 8

**Note: since R = v**^{2} sin ^{2}θ / 8

= v^{2} sin (180 - 2θ) / 8

= v^{2 }sin 2(90 –θ) / 8

Therefore, projectile will possess same range when projected with same velocity making complementary angles, that is, **θ or (90 – θ).**

Trajectory the path followed by the projectile is termed as trajectory. Consider at any time that the particle is at a point **p(x, y)**

That** x = vxt = v cos θ.t or t = x / v cos θ**

At the same instant vertical distance **y = vty – 1 / 2 gt**^{2}

= v sin θ (x / v cosθ) – 8/2 (x / v cos θ)^{2}

= x tan θ - 8x^{2} / 2v^{2} cos θ x tan θ (1 – x / R)

Comparing it with y = ax + bx^{2}

We say that it is general equation of parabola. Hence the path followed by the projectile is a parabola.

Instantaneous velocity **(vt)** let **v** be the velocity at any instant **t**, then

**Vt = v**_{xi} + v’_{yj} = v cos θi + (vsin θ - gt) j

|V'| + √v^{2} + 8(2t2) – 2 vgt sin θ

α = tan -1 (v sin θ - get / v cosθ)

ExpertsMind.com - Physics Assignment Help, Two Dimensions Motion Assignment Help, Two Dimensions Motion Homework Help, Two Dimensions Motion Assignment Tutors, Two Dimensions Motion Solutions, Two Dimensions Motion Answers, Classical Physics Assignment Tutors

**Two Dimensions Motion**

**V = v**

_{xi }+ v_{yj}.Treat the motion in x- and y- directions as one dimensional motion. The acceleration can also be resolved and taken separately along

**x- and y-**directions.

Projectile a freely falling body having constant horizontal velocity may be termed as a projectile. In general, we can say that if velocity along one-direction, that is, x-direction remains constant and there exists along

**y-**direction an accelerated motion, then a body undergoing such a motion is called a projectile. It always follows a parabolic path consider a projectile fired with a velocity v making an angle θ with the horizontal from the point 0 (considered origin) at

**t = 0**. resolving

**v**along

**x –**and

**y-**directions

vx = v cos θ

And vy – v sin θ

vx = v cos θ

And vy – v sin θ

Maximum height attained (hmax) can be calculated by assuming that at maximum height, the vertical component of the velocity become zero.

That is,

Applying

**v**

^{2}– u^{2}= 2as, we get v^{2}y – v^{2}y = - 28h_{max}

Or

**h**

_{max}= v^{2}sin^{2}θ / 2g as v’y = 0Time of flight

**(T)**we calculate the time in which the projectile reaches the highest point, when we double this time it would be the time of flight as it would take same time (whatever it takes to each the highest point) to come back on the ground level at

**Q**.

Applying

**v = u + at**

Let t be the time to reach highest point

**v’y = vy – gt or t = v sin θ/8**

Time of flight

**T = 2t = 2v sin θ/8**

Thus the time of flight is the time taken by the projectile to return to the ground from the instant of projection or it is the time for which the projectile remains in air above the horizontal plane from the point of projection.

Horizontal range

**(R)**the distance covered along the horizontal direction between the point of projection and point of return on the ground is called horizontal range.

**R = vx x time of flight = v cos θ (2v sin θ / 8) = v**

^{2}sin^{2}θ/8**R**

_{max }(maximum range) Rmax will be obtained when

**sin**

^{2}θ = 1That is,

**2θ = 900 or θ = 450 and Rmax = v**

^{2}/ 8**Note: since R = v**

= v

= v

^{2}sin^{2}θ / 8= v

^{2}sin (180 - 2θ) / 8= v

^{2 }sin 2(90 –θ) / 8Therefore, projectile will possess same range when projected with same velocity making complementary angles, that is,

**θ or (90 – θ).**

Trajectory the path followed by the projectile is termed as trajectory. Consider at any time that the particle is at a point

**p(x, y)**

That

**x = vxt = v cos θ.t or t = x / v cos θ**

At the same instant vertical distance

**y = vty – 1 / 2 gt**

^{2}

= v sin θ (x / v cosθ) – 8/2 (x / v cos θ)

= x tan θ - 8x

Comparing it with y = ax + bx

= v sin θ (x / v cosθ) – 8/2 (x / v cos θ)

^{2}= x tan θ - 8x

^{2}/ 2v^{2}cos θ x tan θ (1 – x / R)Comparing it with y = ax + bx

^{2}

We say that it is general equation of parabola. Hence the path followed by the projectile is a parabola.

Instantaneous velocity

**(vt)**let

**v**be the velocity at any instant

**t**, then

**Vt = v**

|V'| + √v

α = tan -1 (v sin θ - get / v cosθ)

_{xi}+ v’_{yj}= v cos θi + (vsin θ - gt) j|V'| + √v

^{2}+ 8(2t2) – 2 vgt sin θα = tan -1 (v sin θ - get / v cosθ)

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