In horizontal projectile motion, a body is projected with an initial velocity [math]u[/math] in the horizontal direction from a height [math]H[/math]. The motion can be analyzed as a combination of two independent components:
- Horizontal motion: Uniform motion with constant velocity [math]u[/math].
- Vertical motion: Uniformly accelerated motion under the influence of gravity [math]g[/math] (taken as negative for downward motion).
Theory:
- The horizontal velocity remains constant because there is no acceleration in the horizontal direction.
- The vertical motion is influenced only by gravity, leading to an acceleration [math]g[/math].
- The trajectory of the body is a parabola, as the horizontal and vertical motions combine to form a curved path.
1. Equations of Motion
Let the body be projected from a height [math]H[/math] with initial velocity [math]u[/math]. The horizontal and vertical components of motion are given as follows:
Horizontal Motion:
- Initial velocity: [math]v_x = u[/math]
- Displacement in horizontal direction:
[math]x = u t[/math]
Vertical Motion:
- Initial velocity: [math]v_y = 0[/math]
- Displacement in vertical direction:
[math]y = -\frac{1}{2} g t^2[/math]
Combining both components, the position vector at any time [math]t[/math] is:
[math]\vec{r} = (u t)\hat{i} + \left(-\frac{1}{2} g t^2\right)\hat{j}[/math]
Cartesian Equation of Trajectory for Horizontal Projectile Motion
To derive the trajectory equation of the horizontal projectile, we eliminate time [math]t[/math] from the horizontal and vertical motion equations.
Horizontal Motion:
The horizontal displacement is given by:
[math]x = u t[/math]
From this, solve for [math]t[/math]:
[math]t = \frac{x}{u}[/math]
Vertical Motion:
The vertical displacement is:
[math]y = -\frac{1}{2} g t^2[/math]
Substitute [math]t = \frac{x}{u}[/math] into the vertical motion equation:
[math]y = -\frac{1}{2} g \left(\frac{x}{u}\right)^2[/math]
Simplify:
[math]y = -\frac{g}{2 u^2} x^2[/math]
Final Trajectory Equation:
The Cartesian equation of the trajectory is:
[math]y = -\frac{g}{2 u^2} x^2[/math]
Key Observations:
- The trajectory is a parabola of the form:
[math]y = a x^2[/math], where [math]a = -\frac{g}{2 u^2}[/math]. - The equation shows that the vertical displacement [math]y[/math] is proportional to the square of the horizontal displacement [math]x[/math].
2. Time of Flight ([math]T[/math])
The time of flight is the time taken by the body to hit the ground. This depends only on the vertical motion.
From vertical displacement:
[math]y = -H[/math]
Substitute [math]y = -\frac{1}{2} g t^2[/math]:
[math]-H = -\frac{1}{2} g T^2[/math]
Rearranging:
[math]T^2 = \frac{2H}{g}[/math]
Taking the square root:
[math]T = \sqrt{\frac{2H}{g}}[/math]
3. Horizontal Range ([math]R[/math])
The horizontal range is the distance covered in the horizontal direction before the body hits the ground.
Using horizontal motion:
[math]x = u t[/math]
For the total flight time [math]T[/math]:
[math]R = u T[/math]
Substitute [math]T = \sqrt{\frac{2H}{g}}[/math]:
[math]R = u \sqrt{\frac{2H}{g}}[/math]
4. Velocity at Any Instant ([math]\vec{v}[/math])
At any time [math]t[/math], the velocity has both horizontal and vertical components:
Horizontal Component:
[math]v_x = u[/math]
Vertical Component:
From vertical motion:
[math]v_y = -g t[/math]
The velocity vector is:
[math]\vec{v} = v_x \hat{i} + v_y \hat{j}[/math]
Substitute the components:
[math]\vec{v} = u \hat{i} + (-g t) \hat{j}[/math]
Magnitude of Velocity:
The magnitude of velocity is given by:
[math]v = \sqrt{v_x^2 + v_y^2}[/math]
Substitute the components:
[math]v = \sqrt{u^2 + (-g t)^2}[/math]
[math]v = \sqrt{u^2 + g^2 t^2}[/math]
5. Angle with Which the Body Hits the Horizontal
The angle [math]\theta[/math] that the velocity vector makes with the horizontal is given by:
[math]\tan\theta = \frac{v_y}{v_x}[/math]
Substitute the components:
[math]\tan\theta = \frac{-g t}{u}[/math]
At the time of hitting the ground ([math]t = T[/math]):
[math]\tan\theta = \frac{-g T}{u}[/math]
Substitute [math]T = \sqrt{\frac{2H}{g}}[/math]:
[math]\tan\theta = \frac{-g \sqrt{\frac{2H}{g}}}{u}[/math]
Simplify:
[math]\tan\theta = \frac{-\sqrt{2gH}}{u}[/math]
The angle [math]\theta[/math] is:
[math]\theta = \tan^{-1}\left(\frac{-\sqrt{2gH}}{u}\right)[/math]
Final Summary:
- Time of Flight:
[math]T = \sqrt{\frac{2H}{g}}[/math]
Horizontal Range:
[math]R = u \sqrt{\frac{2H}{g}}[/math]
Velocity at Any Instant:
[math]\vec{v} = u \hat{i} + (-g t) \hat{j}[/math]
[math]v = \sqrt{u^2 + g^2 t^2}[/math]
Angle of Impact:
[math]\theta = \tan^{-1}\left(\frac{-\sqrt{2gH}}{u}\right)[/math]
