## Saturday, 17 March 2012

### Projectile

If a body is thrown up in the air such that it deviates from simple one dimensional motion, it is termed a projectile. The analysis of motion associated with a projectile is known as projectile motion The path followed by a projectile is called its trajectory, which is directly influenced by gravity.
The initial velocity v0 can be written as

$v_0=v_{0x}\mathbf{i} + v_{0y}\mathbf{j}$
The components v0x and v0y can be found if the angle θ0 is known:
$v_{0x}=v_0\cos\theta_0$
and
$v_{0y}=v_0\sin\theta_0$
The horizontal motion and the motion are independent of each other ; that is, neither motion affects the other.
Since there is no acceleration in the horizontal direction, the horizontal component of the velocity remains unchanged throughout the motion.
The vertical motion is the motion of a particle in free fall. Equations for free fall apply. For example, $y-y_0=(v_0\sin\theta_0)t-\tfrac{1}{2}gt^2$. Other useful equations for the vertical y-axis are $v_y=v_0\sin\theta_0-gt$, and $v_y^2=(v_0\sin\alpha_0)^2-2g(y-y_0)$.
Eliminating t between the following two equations, $x-x_0=(v_0\cos\theta_0)t$ and $y-y_0=(v_0\sin\theta_0)t-\tfrac{1}{2}gt^2$, we obtain the equation of the path (the trajectory) of the projectile:
$y=x\tan(\theta)-\frac{x^2g}{2v^2_{0}cos^2 \theta}$
Time to reach the maximum height
$t={v_{0y}\over g}$
Time to reach ground
$t={2v_{0y}\over g}$
Displacement in X direction
$\Delta x={v_{0x}t}$
Displacement in Y direction
$\Delta y={v_{0y}t-{\tfrac{1}{2}}gt^2}$
Range of projectile
${r=\frac{v^2}{g}\sin(2\theta)}\,$
Maximum height
$y_{max}={{v^2\sin^2\theta}\over 2g}$