The launch of any object whose initial velocity forms an angle (other than \(90^o\) ) with the Earth's surface is called oblique throw. Here, we consider only the case without friction; that is, the air resistance is neglected. We also neglect the variation of gravity in relation to the height. This study is for objects that have such formats that suffer little aerodynamic drag, such as an arrow, and only for movements near the surface of the Earth. This methodology cannot be used to treat the movement of a rocket, for example. In the case of the movement of a rocket, the gravity will change with the altitude and the aerodynamic friction will depend on the shape of the rocket, and it is not possible to ignore such characteristics of the movement.

Oblique launch at vacuum

In this type of launch, the only acceleration come from Earth's gravity.

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Animation of a cannonball being thrown from a tower. The magnitudes are: \(v_0 \) is the initial velocity, \(\alpha_0 \) is the velocity angle with the x-axis and \(y_0 \) is the starting position.

To facilitate the analysis of these types of problems, the movement is separated into two components:

In the horizontal (projection \(x\) )
The object describes a uniform movement.
In the vertical (projection \(y\) )
It describes an object in a uniformly accelerated motion, analogous to the vertical launch in vacuum.
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The figure above shows an oblique throw. If the speed \(v_0\) is decomposed on the axis \(x\), \(v_{0x} = v_0 cos(\alpha_0)\), and the axis \(y\), \(v_{0y} = v_0 sen(\alpha_0)\), the following functions of the positions can be written: \begin{align} y(t) &= y_0 + v_0 sen(\alpha_0) t - g\frac{t^2}{2} \notag \\ x(t) &= x_0 + v_0 cos(\alpha_0) t \notag \end{align} On the other hand, the equations for speeds are: \begin{align} v_y(t) &= v_0 sen(\alpha_0) - gt \notag \\ v_x(t) &= v_0 cos(\alpha_0) \notag \end{align}

In a particular case of oblique release, where the object returns to the same height from which it was launched, for a given initial velocity \(v_0\) at an angle \(\alpha_0\) with the horizontal, the range \(\Lambda\) is $$ \Lambda = \frac{v_0^2 sen(2 \alpha_0)}{g} $$ and the maximum height \(H\) is $$ H = \frac{(v_0 sen(\alpha_0))^2}{2 g}.$$

It is also important to note that:

  • The range is maximum when \(\alpha_0 = 45^o\).
  • Complementary launch angles, i.e., where the sum of the angles are equal to \(90^o\), result in the same range. For example, the launch of an object with an angle of \(30^o\) has the same extent if the launch is made with \(60^o\).
  • The horizontal launch is a particular case of the oblique throw, where an object is launched with an initial horizontal speed from above a building, for example.