Angular Projectile Motion
In angular projectile motion, a body is projected with an initial velocity [math]u[/math] at an angle [math]\theta[/math] with respect to the horizontal. The motion is two-dimensional and follows a parabolic trajectory due to the influence of gravity.
- Horizontal Motion:
- The horizontal velocity remains constant as there is no acceleration in the horizontal direction.
- The displacement in the horizontal direction increases uniformly with time.
- Vertical Motion:
- The vertical velocity decreases due to the effect of gravity.
- At the highest point of the trajectory, the vertical velocity becomes zero.
- The body accelerates downward due to gravity after reaching the highest point.
1. Equation of Motion
The position of the projectile at any time [math]t[/math] is determined by resolving the motion into horizontal and vertical components.
Horizontal Motion:
- Initial velocity in the horizontal direction:
[math]v_x = u \cos\theta[/math] - Displacement in the horizontal direction:
[math]x = (u \cos\theta) t[/math]
Vertical Motion:
- Initial velocity in the vertical direction:
[math]v_y = u \sin\theta[/math] - Displacement in the vertical direction:
[math]y = (u \sin\theta) t – \frac{1}{2} g t^2[/math]
Thus, the position vector of the projectile at any time [math]t[/math] is:
[math]\vec{r} = (u \cos\theta) t \hat{i} + \left[(u \sin\theta) t – \frac{1}{2} g t^2\right] \hat{j}[/math]
Cartesian Form of Equation of Motion in Angular Projection
The Cartesian equation of motion describes the trajectory of a projectile launched at an angle [math]\theta[/math] with an initial velocity [math]u[/math]. The motion is a parabola. Here’s the detailed derivation:
Equations of Motion in Component Form
Horizontal Motion:
- Initial velocity in the horizontal direction:
[math]v_x = u \cos\theta[/math] - Horizontal displacement:
[math]x = (u \cos\theta) t[/math]
Vertical Motion:
- Initial velocity in the vertical direction:
[math]v_y = u \sin\theta[/math] - Vertical displacement under gravity:
[math]y = (u \sin\theta) t – \frac{1}{2} g t^2[/math]
Expressing Time ([math]t[/math]) in Terms of Horizontal Displacement ([math]x[/math]):
From the horizontal motion equation:
[math]x = (u \cos\theta) t[/math]
Solve for [math]t[/math]:
[math]t = \frac{x}{u \cos\theta}[/math]
Substituting [math]t[/math] into the Vertical Motion Equation:
The vertical displacement equation is:
[math]y = (u \sin\theta) t – \frac{1}{2} g t^2[/math]
Substitute [math]t = \frac{x}{u \cos\theta}[/math]:
[math]y = (u \sin\theta) \frac{x}{u \cos\theta} – \frac{1}{2} g \left(\frac{x}{u \cos\theta}\right)^2[/math]
Simplify the terms:
[math]y = x \tan\theta – \frac{g x^2}{2 u^2 \cos^2\theta}[/math]
Final Cartesian Equation of the Trajectory:
[math]y = x \tan\theta – \frac{g x^2}{2 u^2 \cos^2\theta}[/math]
Key Observations:
- The equation is a parabola of the form:
[math]y = ax + bx^2[/math]
where:- [math]a = \tan\theta[/math]
- [math]b = -\frac{g}{2 u^2 \cos^2\theta}[/math]
- Physical Interpretation:
- The term [math]x \tan\theta[/math] represents the linear rise of the projectile.
- The term [math]-\frac{g x^2}{2 u^2 \cos^2\theta}[/math] represents the downward curvature due to gravity.
Let me know if you’d like further refinements!
2. Time of Flight ([math]T[/math])
The time of flight is the total time the projectile remains in the air. At the end of the flight, the vertical displacement is zero ([math]y = 0[/math]).
From the vertical displacement equation:
[math]y = (u \sin\theta) t – \frac{1}{2} g t^2[/math]
Set [math]y = 0[/math]:
[math](u \sin\theta) t – \frac{1}{2} g t^2 = 0[/math]
Factorize:
[math]t \left[(u \sin\theta) – \frac{1}{2} g t\right] = 0[/math]
This gives two solutions:
[math]t = 0[/math] (at the start)
[math]t = \frac{2 u \sin\theta}{g}[/math]
Thus, the total time of flight is:
[math]T = \frac{2 u \sin\theta}{g}[/math]
3. Horizontal Range ([math]R[/math])
The horizontal range is the horizontal distance traveled during the time of flight.
From horizontal motion:
[math]x = (u \cos\theta) t[/math]
For the total flight time [math]T[/math]:
[math]R = (u \cos\theta) T[/math]
Substitute [math]T = \frac{2 u \sin\theta}{g}[/math]:
[math]R = u \cos\theta \cdot \frac{2 u \sin\theta}{g}[/math]
Simplify using [math]\sin 2\theta = 2 \sin\theta \cos\theta[/math]:
[math]R = \frac{u^2 \sin 2\theta}{g}[/math]
Maximum Range:
The range is maximum when [math]\sin 2\theta = 1[/math], i.e., [math]\theta = 45^\circ[/math].
Maximum range:
[math]R_{\text{max}} = \frac{u^2}{g}[/math]
4. Maximum Height ([math]H_{\text{max}}[/math])
The maximum height is reached when the vertical velocity becomes zero ([math]v_y = 0[/math]).
From vertical motion:
[math]v_y = u \sin\theta – g t[/math]
Set [math]v_y = 0[/math]:
[math]t = \frac{u \sin\theta}{g}[/math]
At this time, the vertical displacement is:
[math]y = (u \sin\theta) t – \frac{1}{2} g t^2[/math]
Substitute [math]t = \frac{u \sin\theta}{g}[/math]:
[math]H_{\text{max}} = (u \sin\theta) \cdot \frac{u \sin\theta}{g} – \frac{1}{2} g \left(\frac{u \sin\theta}{g}\right)^2[/math]
Simplify:
[math]H_{\text{max}} = \frac{u^2 \sin^2\theta}{2 g}[/math]
5. Velocity at Any Instant ([math]\vec{v}[/math])
At any time [math]t[/math], the velocity has horizontal and vertical components:
Horizontal Component:
[math]v_x = u \cos\theta[/math] (remains constant)
Vertical Component:
[math]v_y = u \sin\theta – g t[/math]
The velocity vector is:
[math]\vec{v} = (u \cos\theta) \hat{i} + \left[(u \sin\theta – g t)\right] \hat{j}[/math]
Magnitude of Velocity:
[math]v = \sqrt{v_x^2 + v_y^2}[/math]
Substitute the components:
[math]v = \sqrt{(u \cos\theta)^2 + (u \sin\theta – g t)^2}[/math]
6. Angle of Impact
The angle [math]\theta_{\text{impact}}[/math] that the velocity vector makes with the horizontal is given by:
[math]\tan\theta_{\text{impact}} = \frac{v_y}{v_x}[/math]
Substitute the components:
[math]\tan\theta_{\text{impact}} = \frac{u \sin\theta – g t}{u \cos\theta}[/math]
At the time of hitting the ground ([math]t = T[/math]):
[math]\tan\theta_{\text{impact}} = \frac{u \sin\theta – g \cdot \frac{2 u \sin\theta}{g}}{u \cos\theta}[/math]
Simplify:
[math]\tan\theta_{\text{impact}} = \frac{-u \sin\theta}{u \cos\theta}[/math]
[math]\theta_{\text{impact}} = \tan^{-1}(-\tan\theta)[/math]
Thus, the body hits the ground at an angle equal to [math]\theta[/math] but below the horizontal.
Final Summary:
- Time of Flight:
[math]T = \frac{2 u \sin\theta}{g}[/math]
Horizontal Range:
[math]R = \frac{u^2 \sin 2\theta}{g}[/math]
[math]R_{\text{max}} = \frac{u^2}{g}[/math] (when [math]\theta = 45^\circ[/math])
Maximum Height:
[math]H_{\text{max}} = \frac{u^2 \sin^2\theta}{2 g}[/math]
Velocity at Any Instant:
[math]\vec{v} = (u \cos\theta) \hat{i} + \left[(u \sin\theta – g t)\right] \hat{j}[/math]
[math]v = \sqrt{(u \cos\theta)^2 + (u \sin\theta – g t)^2}[/math]
Angle of Impact:
[math]\theta_{\text{impact}} = \tan^{-1}(-\tan\theta)[/math]
Prove that the horizontal range remains the same when the angle of projection is [math]\theta[/math] or [math](90^\circ – \theta)[/math]:
Horizontal Range Formula
The horizontal range [math]R[/math] for a projectile is given by:
[math]R = \frac{u^2 \sin 2\theta}{g}[/math]
Now, let’s substitute [math]\theta[/math] with [math](90^\circ – \theta)[/math] and show that the formula remains the same.
Substituting [math](90^\circ – \theta)[/math]:
Replace [math]\theta[/math] in the formula with [math](90^\circ – \theta)[/math]:
[math]R’ = \frac{u^2 \sin 2(90^\circ – \theta)}{g}[/math]
Simplify the angle inside the sine function:
[math]2(90^\circ – \theta) = 180^\circ – 2\theta[/math]
Thus:
[math]R’ = \frac{u^2 \sin(180^\circ – 2\theta)}{g}[/math]
Using the Trigonometric Identity:
The sine function satisfies the identity:
[math]\sin(180^\circ – x) = \sin x[/math]
Therefore:
[math]\sin(180^\circ – 2\theta) = \sin 2\theta[/math]
Substitute back into the equation for [math]R'[/math]:
[math]R’ = \frac{u^2 \sin 2\theta}{g}[/math]
Conclusion:
[math]R’ = R[/math]
This proves that the range remains the same for angles [math]\theta[/math] and [math](90^\circ – \theta)[/math].
