Projectile Motion

Projectile motion possesses many interesting properties. One of them is the fact that two projectiles, when thrown with the same speed but at angles \(\theta\) and \(90-\theta\) respectively, land at the same position! You can verify this fact by using the interactive simulation of projectile motion given below. Also, try to hit that blue target!




Description

Basic concepts: \[ x = u t \cos \theta \] \[ y = ut \sin \theta - \frac{1}{2} g t^2 \]
  1. In projectile motion, acceleration due to gravity acts in the vertical direction. Hence, only the y-component of velocity is affected by acceleration due to gravity. The x-component of velocity remains constant. Observe that the value of \(v_x\) remains constant in the above simulation
  2. Let the coordinates of the projectile be denoted by \((x, y)\). Let the initial speed be \( u = \sqrt{u_x^2 + u_y^2} \). Let \( \theta \) be the angle that the vector \(\vec{u}\) makes with the X-axis. The above two equations allow us to determine the position of the projectile at any time t.
  3. During ascent, \(v_y\) is positive. During descent \(v_y\) is negative. At the highest point, \(v_y\) is zero!
Time of ascent, descent and flight: \[ t_f = \frac{2u}{g} \] \[ t_a = t_d = \frac{u}{g} \]
  1. Using the fact that \(v_y\) is zero at the highest point, and using \(v = u - gt\) in the y-direction, we obtain time of ascent \(= \frac{u \sin \theta}{g} \).
  2. If air resistance is ignored, then time of ascent = time of descent. Hence, time of flight = time of ascent + time of descent \( = \frac{2 u \sin \theta}{g} \)
Maximum height: \[ H_{max} = \frac{u^2 \sin \theta}{2g} \]
  1. Using the fact that \(v_y\) is zero at the highest point, and using \( v^2 = u^2 - 2gy \) in the y-direction, we get maximum height \( H = \frac{u^2 sin^2 \theta}{2g} \)
  2. For a given initial velocity \(u\), maximum possible height is achieved when angle of projection is 90 degrees, as is evident from the above formula
Horizontal range: \[ R = \frac{u^2 \sin 2\theta}{g} \]
  1. We know that \(x = u t \cos \theta\). If we substitute \( t = \frac{2 u \sin \theta}{g} \) (time of flight) and we use \( 2 \sin \theta \cos \theta = \sin 2 \theta \), then we get \( R = \frac{u^2 \sin 2 \theta}{g} \)
  2. As can be deciphered from the above formula, range is maximum when \( \theta = 45^{\circ} \)
  3. You can verify here that \( \sin(180 - A) = \sin \theta \). Hence, the range is same for angles of projection of \(\theta\) and \(90 - \theta\), for the same initial velocity \(u\), since \( \sin (180 - 2\theta) = \sin 2 \theta \).


Notes:
  1. The above simulation is fully compatible with SI units. Acceleration due to gravity = 9.81 acceleration units
  2. The length of the canvas corresponds to (maximum possible range + 40) length units. Maximum range is achieved when θ = 45 degrees and u = 100 velocity units
  3. Related: Spring Motion, Planetary Motion, Double Pendulum.

Developed by ChanRT | Fork me at GitHub