## 2-D motion with gravity

When an object (projectile) is thrown near the surface of the earth, the only force acting on it is gravity. The horizontal and vertical motions are independent. The initial horizontal velocity (\(v_x\)) remains constant because there is no force in that direction. The velocity in the vertical direction (\(v_y\)) is altered by the force of gravity over time, \(t\): \(v_y' = v_y - g t \), where \(g\) is the acceleration due to gravity. For earth, \(g = 9.81 m/s^2\) and depends on the gravitational constant \(G = 6.67×10^{−11}\,N·(m/kg)^2\), the mass of the earth, \(m_e\) and the distance from the center of the earth \(r_e\): $$g = G \frac{m_e}{r_e^2}$$ (part of Newton's law of gravitation \(F=G{\frac {m_{1}m_{2}}{r^{2}}}\ \) describing the gravitational force between two masses and the distance \(r \) between them)

Projectiles are a great way to learn about 2-D motion in the presence of gravity

The basic equations for velocity and position are as follows:

### 2-D projectiles

Enter values for the initial height (≥ 0), velocity, launch angle (θ), range, and acceleration due to gravity (g = 9.81 m/s^{2} for earth). Air resistance is neglected.

Four of these quantities are sufficient to calculate the fifth. Select the radio button for the one you want to calculate.

Quantities entered must be units of meters, m/s and degrees.

Some combinations of inputs are invalid, which is the case if you see "NaN" (not a number), or the numbers reset to something that is physically possible.

Velocity in the y-direction vs time:

#### More projectile equations

For an object with a starting height of \(y_0\), the range is maximized by the angle
$$\theta = \arccos \sqrt{ \frac {2 g y_0 + v^2} {2 g y_0 + 2v^2}} $$
This reduces to \(\arccos( 1/\sqrt{2}) = 45^{\circ}\) for \(y_0=0\). In that instance, range is given by \(R = v^2/g\) and the time of flight is \(t=\sqrt{2}\,v/g\).

The general time of flight, for any starting height, is given by $$t_g = \frac {v \sin \theta} {g} + \frac {\sqrt{v^2 \sin^2 \theta + 2 g y_0}} {g}$$

The maximum height of a projectile is $$h_{max} = \frac{v_0^2 \sin^2(\theta)}{2g}$$ The maximum time aloft and the maximum height are achieved by setting the launch angle to 90°.

The maximum range for a projectile starting at \(y_0 = 0\) is $$d = \frac{v_0^2}{g}\sin(2\theta)$$ and the general equation for any starting height is

$$d = \frac{v \cos \theta}{g} \left( v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0} \right) $$