Understanding Time of Flight in Projectile Motion

The time of flight of a projectile is the total time duration for which the projectile remains in the air from the moment it is launched until it hits the ground again. For a projectile launched from a horizontal plane and landing back on the same horizontal plane, we can determine the time of flight by considering its vertical motion.

Analyzing Vertical Motion

Let's consider the initial velocity u of the projectile and the angle of inclination θ with the horizontal. The gravitational acceleration is denoted by g.

The initial velocity can be resolved into horizontal and vertical components:

• Horizontal component of velocity, \(u_x = u \cos \theta\)
• Vertical component of velocity, \(u_y = u \sin \theta\)

The vertical motion is affected by gravity, which acts downwards. We can use the kinematic equation relating displacement, initial velocity, time, and acceleration:

\(s = u_y t + \frac{1}{2} a_y t^2\)

Here:

• \(s\) is the vertical displacement.
• \(u_y\) is the initial vertical velocity.
• \(a_y\) is the vertical acceleration (due to gravity).
• \(t\) is the time.

Deriving the Time of Flight Formula

When the projectile is launched from a horizontal plane and lands back on the same plane, the net vertical displacement (\(s\)) is zero.

The vertical acceleration \(a_y\) is due to gravity, acting downwards. Taking the upward direction as positive, \(a_y = -g\).

Substituting \(s = 0\), \(u_y = u \sin \theta\), and \(a_y = -g\) into the kinematic equation:

\(0 = (u \sin \theta) T + \frac{1}{2} (-g) T^2\)

Where \(T\) represents the total time of flight.

We can rearrange the equation:

\(0 = T (u \sin \theta - \frac{1}{2} g T)\)

This equation has two possible solutions for \(T\):

1. \(T = 0\): This corresponds to the initial moment of launch.
2. \(u \sin \theta - \frac{1}{2} g T = 0\): This corresponds to the moment the projectile lands back on the horizontal plane.

Solving the second equation for \(T\):

\(u \sin \theta = \frac{1}{2} g T\)

\(T = \frac{2 u \sin \theta}{g}\)

This formula gives the time of flight of a projectile launched from a horizontal plane with initial velocity u at an angle θ with the horizontal, under gravitational acceleration g.

Conclusion

The time of flight of a projectile on a horizontal plane is given by the formula \(\frac{{2{\rm{u}}\sin \theta }}{{\rm{g}}}\), where u is the initial velocity, θ is the angle of inclination, and g is the gravitational acceleration.