# Orbit and Maximum Height

If the projectile is launched with an initial velocity v0, then it can be written as

\mathbf{v}_0 = v_{0x}\mathbf{i} + v_{0y}\mathbf{j}.

The components v0x and v0y can be found if the angle, ϴ is known:

v_{0x} = vgh_0\cos\theta,

v_{0y} = v_0\sin\theta.

If the projectile's range, launch angle, and drop height are known, launch velocity can be found by

V_0 = \sqrt{{R^2 g} \over {R \sin 2\theta + 2h \cos^2\theta}} .

The launch angle is usually expressed by the symbol theta, but often the symbol alpha is used. Kinematic quantities of projectile motion

In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other. Acceleration

Since there is no acceleration in the horizontal direction velocity in horizontal direction is constant which is equal to ucosα. The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, equal to g.[1] The components of the acceleration:

a_x = 0 ,

a_y = -g .

Velocity

The horizontal component of the velocity remains unchanged throughout the motion. The vertical component of the velocity increases linearly, because the acceleration is constant. At any time t, the components of the velocity:

v_x=v_0 \cos(\theta) ,

v_y=v_0 \sin(\theta) - gt .

The magnitude of the velocity (under the Pythagorean theorem):

v=\sqrt{v_x^2 + v_y^2 \ } .

Displacement

Displacement and coordinates of parabolic throwing

At any time t, the projectile's...

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