Revision/fast recall
1. TITLE
Projectile Motion cheat sheet — Everything for NEET on one page
2. PROJECTILE MOTION — QUICK FACTS
- Projectile motion combines independent horizontal and vertical motions.
- Horizontal acceleration ($a_x$) is always zero.
- Vertical acceleration ($a_y$) is constantly $-g$ (downward).
- Horizontal velocity ($v_x$) remains completely constant throughout flight.
- The path (trajectory) of a projectile is always a parabola.
- At maximum height, vertical velocity is exactly zero.
- At maximum height, velocity is purely horizontal ($v_{ox}\hat{i}$).
- Time of flight ($T_f$) is twice the time to reach maximum height ($t_m$).
- Maximum range occurs at a launch angle of $45^\circ$.
- Ranges are identical for angles that are equally above/below $45^\circ$.
3. FORMULAS:
Time of Flight ($T_f$): $T_f = \frac{2v_0 \sin\theta_0}{g}$ [Unit: s]
→ Use when: Calculating the total time the projectile remains in the air.
Time to Maximum Height ($t_m$): $t_m = \frac{v_0 \sin\theta_0}{g}$ [Unit: s]
→ Use when: Finding the exact time the projectile reaches its peak.
Maximum Height ($h_m$): $h_m = \frac{v_0^2 \sin^2\theta_0}{2g}$ [Unit: m]
→ Use when: Determining the highest vertical point reached.
Horizontal Range ($R$): $R = \frac{v_0^2 \sin 2\theta_0}{g}$ [Unit: m]
→ Use when: Finding the total horizontal distance covered.
Equation of Trajectory: $y = x \tan\theta_0 - \frac{g x^2}{2v_0^2 \cos^2\theta_0}$ [Unit: m]
→ Use when: Relating $x$ and $y$ coordinates directly without knowing the time.
4. KEY VALUES:
Acceleration due to gravity ($g$) = $9.8 \text{ m s}^{-2}$ (often approx. $10 \text{ m s}^{-2}$)
Optimal angle for Maximum Range ($\theta_0$) = $45^\circ$
Height at Maximum Range ($h_m$ at $45^\circ$) = $\frac{R_{max}}{4}$
5. COMPARISON SUMMARY
[Motion Component][Horizontal Axis (x)][Vertical Axis (y)]
[Acceleration][$a_x = 0$][ $a_y = -g$]
[Velocity][$v_x = v_0 \cos\theta_0$ (Constant)][$v_y = v_0 \sin\theta_0 - gt$ (Changes)]
[Nature of Motion][Uniform Velocity][Uniformly Accelerated]
6. COMMON MISTAKES — QUICK LIST
DON'T:
* Don't assume the total velocity is zero at maximum height (horizontal velocity still exists!).
* Don't mix up the formulas for Time of Flight ($T_f$) and Time of Maximum Height ($t_m$).
* Don't use the standard Range formula if the starting and landing heights are different.
DO:
* Do remember to treat 2D projectile motion simply as two independent 1D motions.
* Do exploit symmetry: Time up equals time down ($T_f = 2t_m$).
7. MNEMONICS & MEMORY TRICKS
REMEMBER:
"Height has squares everywhere": Helps remember the Maximum Height ($h_m$) formula has $v_0^2$, $\sin^2\theta_0$, AND $2g$ in the denominator.
"Range has a Two inside": Helps remember the Range ($R$) formula uses $\sin(2\theta_0)$ and a plain $g$ in the denominator.
"Complementary Angles, Identical Ranges"
8. HIGH-YIELD NEET POINTS
NEET LOVES TO ASK:
* Velocity at Peak: At the absolute peak of the trajectory, acceleration is $-g$ and velocity is purely horizontal ($v_0 \cos\theta_0$).
* Equal Ranges: The horizontal range is identical for complementary projection angles like $\theta$ and $90^\circ - \theta$ (or $45^\circ + \alpha$ and $45^\circ - \alpha$).
* Max Range to Height relation: When a projectile is thrown for maximum horizontal range ($45^\circ$), the maximum height achieved is exactly one-quarter of that range ($h_m = R_{max} / 4$).
9. ONE MCQ — SELF-TEST
If a projectile is launched such that its maximum horizontal range ($R_{max}$) is $100\text{ m}$, what is the maximum height ($h_m$) it can achieve if launched at that exact same optimal angle?
(a) $25\text{ m}$
(b) $50\text{ m}$
(c) $100\text{ m}$
(d) $200\text{ m}$
Answer: (a) $25\text{ m}$. For maximum range, the angle is $45^\circ$. At $45^\circ$, $h_m = R_{max}/4 = 100/4 = 25\text{ m}$.
10. NCERT PAGE REFERENCES
Revise from:
- Page 39: Equation of path of a projectile
- Page 39: Time of maximum height
- Page 40: Maximum height and horizontal range
11. CTA
Detailed explanation → Kinematics Guide
Quick test → Mock exams