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Equation of a Trajectory Simplified Revision Notes

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2.6.3 Equation of a Trajectory

infoNote

The equation of trajectory describes the path of an object under projectile motion, typically in a parabolic form. It combines both horizontal and vertical components of motion, assuming constant acceleration due to gravity and no air resistance.

infoNote

For an object launched with initial velocity uu at an angle θ\theta from the horizontal:

  • Horizontal displacement ( xx ): x=ucos(θ)tx = u \cos(\theta) \cdot t
  • Vertical displacement ( yy ): y=usin(θ)t12gt2y = u \sin(\theta) \cdot t - \frac{1}{2} g t^2 By eliminating time tt , the trajectory equation is:
y=xtan(θ)gx22u2cos2(θ)y = x \tan(\theta) - \frac{g x^2}{2u^2 \cos^2(\theta)}

This equation represents a parabola and shows the relationship between horizontal and vertical displacements of the projectile.

Projectile: Cartesian Equation of Motion

infoNote

Example:

A particle is projected at an angle of 6060^\circ above the horizontal with an initial speed of 40ms40 ms. Find the Cartesian equation for the path (trajectory) of the particle.


Steps:

1. Note all initial conditions as vectors in SUVAT.

Given:

  • u=(40cos60,40sin60)\mathbf{u} = \left(40\cos60^\circ, 40\sin60^\circ\right)
  • a=(0,9.8)\mathbf{a} = \left(0, -9.8\right)
  • s=?\mathbf{s} = ?
  • t=tt = t

2. Use s=ut+12at2\mathbf{s} = \mathbf{ut} + \frac{1}{2} \mathbf{a}t^2 to get the parametric equations of motion in terms of t.t.

s=ut+12at2\mathbf{s} = \mathbf{ut} + \frac{1}{2} \mathbf{a}t^2(xy)=((40cos60)t(40sin60)t)+12(09.8)t2\left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} (40\cos60^\circ)t \\ (40\sin60^\circ)t \end{array} \right) + \frac{1}{2} \left( \begin{array}{c} 0 \\ -9.8 \end{array} \right)t^2(xy)=(20t203t)+(04.9t2)\left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} 20t \\ 20\sqrt{3}t \end{array} \right) + \left( \begin{array}{c} 0 \\ -4.9t^2 \end{array} \right)(xy)=(20t203t4.9t2)\left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} 20t \\ 20\sqrt{3}t - 4.9t^2 \end{array} \right)

3. Write the equation in xx and yy in terms of t, then eliminate t.

Given:

x=20tx = 20t

y=203t4.9t2y = 20\sqrt{3}t - 4.9t^2

Substituting t=x20t = \frac{x}{20} into the equation for yy:

y=203(x20)4.9(x20)2y = 20\sqrt{3}\left(\frac{x}{20}\right) - 4.9\left(\frac{x}{20}\right)^2y=3x4.9(x2400)y = \sqrt{3}x - 4.9\left(\frac{x^2}{400}\right)y=3x0.01225x2y = \sqrt{3}x - 0.01225x^2


Projectiles: Horizontal Projectiles

When an object is projected horizontally, after projection, the only force acting on the object is its weight. The initial velocity is always of the form (u,0)ms1(u, 0)ms^{-1}, where u is the initial horizontal speed.

image
infoNote

Example:

A particle is projected horizontally at an initial speed of 15ms115 ms^{-1}. 10 seconds later, it hits the ground. Find the distance above the ground from which it was projected.

Solution:

Given:

  • Initial velocity u=(15,0)ms\mathbf{u} = (15, 0)ms
  • Acceleration a=(0,9.8)ms2\mathbf{a} = (0, -9.8) ms^2
  • Time t=10st = 10 s
  • Displacement s=?\mathbf{s} = ?

Steps:

  1. Set positive directions:
  • Positive xx-direction is rightward (horizontal)
  • Positive yy-direction is upward (vertical)
  1. Use the SUVAT equation:
s=ut+12at2\mathbf{s} = \mathbf{ut} + \frac{1}{2}\mathbf{a}t^2
  1. Calculate the horizontal displacement x:
x=15×10=150 mx = 15 \times 10 = 150 \text{ m}
  1. Calculate the vertical displacement h:
y=0×10+12(9.8)(10)2=490 my = 0 \times 10 + \frac{1}{2}(-9.8)(10)^2 = -490 \text{ m}

Therefore, the height h=490mh = 490 m.

Note:

To calculate the total distance from the origin O, use:

Total Distance=1502+4902=512 m (3 significant figures)\text{Total Distance} = \sqrt{150^2 + 490^2} = 512 \text{ m (3 significant figures)}

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