Concept Card: Oblique Projectile Motion
1. Physical Framework & Initial Conditions:
A particle launched from ground level with initial speed $u$ at an angle of elevation $\theta$ ($0 < \theta < 90^\circ$) to the horizontal moves in a vertical plane under the constant downward acceleration of gravity $\vec{g}$.
Taking the launch point as origin $O(0, 0)$ with horizontal as $+x$ and vertically upward as $+y$:
- Initial Velocity Components: $u_x = u\cos\theta, \quad u_y = u\sin\theta$
- Acceleration Components: $a_x = 0, \quad a_y = -g$
- Instantaneous Velocity:
$v_x(t) = u\cos\theta \quad (\text{constant throughout flight})$
$v_y(t) = u\sin\theta – gt$
Speed: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 – 2gy}$
Angle with horizontal: $\tan\beta = \frac{v_y}{v_x} = \frac{u\sin\theta – gt}{u\cos\theta}$
2. Core Kinematic Formulas:
- Time of Flight ($T$):
Vertical displacement upon landing is $y = 0$:
$0 = u\sin\theta \cdot T – \frac{1}{2}g T^2 \implies T = \frac{2u\sin\theta}{g} = \frac{2u_y}{g}$
Time of ascent = Time of descent: $t_a = t_d = \frac{u\sin\theta}{g} = \frac{T}{2}$. - Maximum Height Reached ($H_{\text{max}}$):
At the apex of trajectory, vertical velocity vanishes ($v_y = 0$):
$0 = u_y^2 – 2g H_{\text{max}} \implies H_{\text{max}} = \frac{u^2\sin^2\theta}{2g} = \frac{u_y^2}{2g}$
Velocity at top is purely horizontal: $v_{\text{top}} = u_x = u\cos\theta$.
Kinetic energy at top: $K_{\text{top}} = \frac{1}{2}m v_{\text{top}}^2 = \frac{1}{2}mu^2\cos^2\theta = K_0\cos^2\theta$. - Horizontal Range ($R$):
$R = u_x \cdot T = (u\cos\theta)\left(\frac{2u\sin\theta}{g}\right) = \frac{u^2\sin(2\theta)}{g} = \frac{2u_x u_y}{g}$- Maximum range occurs at $\theta = 45^\circ$: $R_{\text{max}} = \frac{u^2}{g}$.
- At $\theta = 45^\circ$, the maximum height is $H = \frac{R_{\text{max}}}{4}$.
- Fundamental Range-Height Identity:
$\frac{H}{R} = \frac{u^2\sin^2\theta / (2g)}{u^2(2\sin\theta\cos\theta) / g} = \frac{\tan\theta}{4} \implies R\tan\theta = 4H \iff \tan\theta = \frac{4H}{R}$
3. Complementary Angles of Projection ($\theta$ and $90^\circ – \theta$):
For a given launch speed $u$, the horizontal range is identical for complementary launch angles ($\theta$ and $90^\circ – \theta$):
- $R_1 = R_2 = R = \frac{u^2\sin(2\theta)}{g}$
- Product of flight times: $T_1 T_2 = \left(\frac{2u\sin\theta}{g}\right)\left(\frac{2u\cos\theta}{g}\right) = \frac{2R}{g} \implies R = \frac{1}{2}g T_1 T_2$
- Geometric mean of maximum heights: $H_1 H_2 = \frac{R^2}{16} \implies R = 4\sqrt{H_1 H_2}$
- Sum of maximum heights: $H_1 + H_2 = \frac{u^2}{2g}$
4. Equation of Trajectory (Parabolic Path):
Eliminating $t = \frac{x}{u\cos\theta}$ yields the standard trajectory equation:
$y = x\tan\theta – \frac{g x^2}{2u^2\cos^2\theta}$
Factored Form (Extremely High-Yield for JEE):
$y = x\tan\theta\left(1 – \frac{x}{R}\right)$
5. Radius of Curvature ($R_c$):
- At the highest point (top): $v = u\cos\theta$ and $\vec{g} \perp \vec{v} \implies a_n = g$:
$R_{c,\text{top}} = \frac{v_{\text{top}}^2}{g} = \frac{u^2\cos^2\theta}{g}$ - At the launch point ($t = 0$): $v = u$ and $a_n = g\cos\theta$:
$R_{c0} = \frac{u^2}{g\cos\theta}$
Solved Examples
Example 1 (Comprehensive Parameters & Kinetic Energy Ratio):
A projectile of mass $m = 0.2\text{ kg}$ is fired from the ground with speed $u = 50\text{ m/s}$ at an angle $\theta = 37^\circ$ to the horizontal ($\cos 37^\circ = 0.8, \sin 37^\circ = 0.6$). Taking $g = 10\text{ m/s}^2$:
(a) Find the time of flight $T$, maximum height $H_{\text{max}}$, and horizontal range $R$.
(b) Find the velocity vector and speed at $t = 2.0\text{ s}$.
(c) Find the kinetic energy of the projectile at the launch point and at the apex.
Solution:
Initial components: $u_x = 50(0.8) = 40\text{ m/s}, \quad u_y = 50(0.6) = 30\text{ m/s}$.
(a) Basic parameters:
– $T = \frac{2u_y}{g} = \frac{2(30)}{10} = 6.0\text{ seconds}$.
– $H_{\text{max}} = \frac{u_y^2}{2g} = \frac{30^2}{2(10)} = \frac{900}{20} = 45\text{ meters}$.
– $R = u_x \cdot T = 40 \times 6.0 = 240\text{ meters}$.
(b) Velocity at $t = 2.0\text{ s}$:
$v_x = u_x = 40\text{ m/s}$
$v_y = u_y – gt = 30 – 10(2.0) = +10\text{ m/s}$
$\vec{v}(2) = 40\hat{i} + 10\hat{j}\text{ m/s}$.
Speed: $v = \sqrt{40^2 + 10^2} = \sqrt{1600 + 100} = \sqrt{1700} = 10\sqrt{17}\text{ m/s} \approx 41.23\text{ m/s}$.
(c) Kinetic energy:
At launch: $K_0 = \frac{1}{2}m u^2 = \frac{1}{2}(0.2)(50^2) = 0.1 \times 2500 = 250\text{ Joules}$.
At apex ($v_{\text{top}} = u_x = 40\text{ m/s}$):
$K_{\text{top}} = \frac{1}{2}m v_x^2 = \frac{1}{2}(0.2)(40^2) = 0.1 \times 1600 = 160\text{ Joules}$.
(Notice: $K_{\text{top}} = K_0\cos^2(37^\circ) = 250 \times (0.8)^2 = 250 \times 0.64 = 160\text{ J}$).
Example 2 (Factored Trajectory Equation Clearing a Wall):
A ball thrown from ground level clears a vertical wall of height $h = 6.0\text{ m}$ located at a horizontal distance $d_1 = 4.0\text{ m}$ from the thrower, and lands on the ground at a distance $d_2 = 12.0\text{ m}$ beyond the wall. Find:
(a) The angle of projection $\theta$.
(b) The initial speed of projection $u$ (take $g = 10\text{ m/s}^2$).
Solution:
(a) The total horizontal range is the sum of distances before and after the wall:
$R = d_1 + d_2 = 4.0\text{ m} + 12.0\text{ m} = 16.0\text{ meters}$.
The wall’s coordinates are $x = 4.0\text{ m}, y = 6.0\text{ m}$.
Applying the factored trajectory equation $y = x\tan\theta\left(1 – \frac{x}{R}\right)$:
$6.0 = 4.0\tan\theta\left(1 – \frac{4.0}{16.0}\right) = 4.0\tan\theta\left(1 – \frac{1}{4}\right) = 4.0\tan\theta\left(\frac{3}{4}\right) = 3.0\tan\theta$.
$\tan\theta = \frac{6.0}{3.0} = 2 \implies \theta = \arctan(2) \approx 63.43^\circ$.
(b) From $\tan\theta = 2$, we have $\sin\theta = \frac{2}{\sqrt{5}}$ and $\cos\theta = \frac{1}{\sqrt{5}}$.
Using the range formula: $R = \frac{2u^2\sin\theta\cos\theta}{g}$:
$16 = \frac{2u^2 \left(\frac{2}{\sqrt{5}}\right)\left(\frac{1}{\sqrt{5}}\right)}{10} = \frac{4u^2}{50} = \frac{2u^2}{25}$.
$2u^2 = 16 \times 25 = 400 \implies u^2 = 200 \implies u = \sqrt{200} = 10\sqrt{2}\text{ m/s} \approx 14.14\text{ m/s}$.
Example 3 (Complementary Angles & Range Synthesis):
Two projectiles are launched from the same point on a horizontal plane with identical initial speeds $u$ at complementary angles $\theta_1$ and $\theta_2 = 90^\circ – \theta_1$. Their respective maximum heights are measured as $H_1 = 45\text{ m}$ and $H_2 = 20\text{ m}$. Taking $g = 10\text{ m/s}^2$:
(a) Calculate the common horizontal range $R$.
(b) Determine the initial launch speed $u$.
(c) Find the times of flight $T_1$ and $T_2$ and verify $R = \frac{1}{2}g T_1 T_2$.
Solution:
(a) Using the complementary height-range theorem:
$R = 4\sqrt{H_1 H_2} = 4\sqrt{45 \times 20} = 4\sqrt{900} = 4 \times 30 = 120\text{ meters}$.
(b) From the sum of maximum heights for complementary angles:
$H_1 + H_2 = \frac{u^2}{2g} \implies 45 + 20 = \frac{u^2}{2(10)} \implies 65 = \frac{u^2}{20} \implies u^2 = 1300$.
$u = \sqrt{1300} = 10\sqrt{13}\text{ m/s} \approx 36.06\text{ m/s}$.
(c) Times of flight from maximum heights ($H = \frac{1}{8}g T^2 \implies T = \sqrt{\frac{8H}{g}}$):
$T_1 = \sqrt{\frac{8 \times 45}{10}} = \sqrt{36} = 6.0\text{ s}$.
$T_2 = \sqrt{\frac{8 \times 20}{10}} = \sqrt{16} = 4.0\text{ s}$.
Checking range product relation:
$R = \frac{1}{2}g T_1 T_2 = \frac{1}{2}(10)(6.0)(4.0) = 5 \times 24 = 120\text{ meters}$ (perfect consistency!).
Example 4 (Perpendicularity of Velocity Vector to Initial Velocity):
A particle is projected from the ground with speed $u$ at an angle $\theta$ to the horizontal.
(a) Derive the expression for the time $t$ when the velocity vector $\vec{v}(t)$ becomes perpendicular to the initial velocity vector $\vec{u}$.
(b) Determine the condition on $\theta$ for this perpendicularity to occur before the particle lands ($t \le T$).
(c) Find the speed of the particle at this instant.
Solution:
(a) Initial velocity: $\vec{u} = u\cos\theta\hat{i} + u\sin\theta\hat{j}$.
Velocity at time $t$: $\vec{v}(t) = u\cos\theta\hat{i} + (u\sin\theta – gt)\hat{j}$.
The vectors are perpendicular when $\vec{u} \cdot \vec{v} = 0$:
$(u\cos\theta)(u\cos\theta) + (u\sin\theta)(u\sin\theta – gt) = 0$
$u^2\cos^2\theta + u^2\sin^2\theta – ugt\sin\theta = 0$
$u^2 – ugt\sin\theta = 0 \implies u = gt\sin\theta \implies t = \frac{u}{g\sin\theta}$.
(b) For this instant to occur during flight ($t \le T = \frac{2u\sin\theta}{g}$):
$\frac{u}{g\sin\theta} \le \frac{2u\sin\theta}{g} \implies 1 \le 2\sin^2\theta \implies \sin^2\theta \ge \frac{1}{2} \implies \sin\theta \ge \frac{1}{\sqrt{2}} \implies \theta \ge 45^\circ$.
Perpendicularity occurs during flight if and only if $\theta \ge 45^\circ$.
(c) Speed at this instant:
Since $\vec{u} \perp \vec{v}$, in the right-angled vector triangle formed by $\vec{u}, \vec{v}$, and $\vec{g}t$ (where $\vec{v} = \vec{u} + \vec{g}t$):
$v = u\cot\theta = \frac{u}{\tan\theta}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A projectile is thrown with kinetic energy $K_0$ at an angle of $60^\circ$ to the horizontal. Its kinetic energy at the highest point of its trajectory is:
(A) $\frac{K_0}{2}$
(B) $\frac{K_0}{4}$
(C) $\frac{3K_0}{4}$
(D) Zero
Problem 2 (JEE Main – Single Correct):
The horizontal range of a projectile is $4\sqrt{3}$ times its maximum height ($R = 4\sqrt{3}H$). The angle of projection $\theta$ with the horizontal is:
(A) $30^\circ$
(B) $45^\circ$
(C) $60^\circ$
(D) $75^\circ$
Problem 3 (JEE Main – Single Correct):
For what angle of projection is the horizontal range equal to the maximum height ($R = H$)?
(A) $\arctan(1) = 45^\circ$
(B) $\arctan(2)$
(C) $\arctan(4)$
(D) $\arctan(0.25)$
Problem 4 (JEE Main – Single Correct):
Two bodies are projected with the same initial speed at angles $\theta$ and $(90^\circ – \theta)$ to the horizontal. If their maximum heights are $H_1$ and $H_2$, their horizontal range $R$ is related to $H_1$ and $H_2$ by:
(A) $R = 4\sqrt{H_1 H_2}$
(B) $R = 2\sqrt{H_1 H_2}$
(C) $R = 4(H_1 + H_2)$
(D) $R = \sqrt{H_1 H_2}$
Problem 5 (JEE Main – Single Correct):
The radius of curvature of a projectile of launch speed $u$ and angle $\theta$ at the highest point of its path is:
(A) $\frac{u^2\cos^2\theta}{g}$
(B) $\frac{u^2\sin^2\theta}{g}$
(C) $\frac{u^2}{g\cos\theta}$
(D) $\frac{u^2}{g}$
Problem 6 (JEE Advanced – One or More Correct):
A projectile launched from the origin has trajectory equation $y = \sqrt{3}x – 2x^2$ (with $x, y$ in meters). Taking $g = 10\text{ m/s}^2$, which of the following statements is/are correct?
(A) The angle of projection is $60^\circ$.
(B) The horizontal range of the projectile is $\frac{\sqrt{3}}{2}\text{ m}$.
(C) The initial speed of projection is $\sqrt{10}\text{ m/s}$.
(D) The maximum height reached is $\frac{3}{8}\text{ m}$.
Problem 7 (JEE Advanced – One or More Correct):
For an oblique projectile launched with speed $u$ at angle $\theta$ to the horizontal over level ground:
(A) The change in momentum between launch and landing is $2mu\sin\theta$ directed vertically downwards.
(B) The velocity vector becomes perpendicular to the initial velocity vector at $t = \frac{u}{g\sin\theta}$.
(C) The perpendicularity of $\vec{v}(t)$ with $\vec{u}$ is physically achievable during flight only if $\theta \ge 45^\circ$.
(D) The average velocity over the entire flight is $u\cos\theta\hat{i}$.
Problem 8 (JEE Advanced – One or More Correct):
A ball is projected from the ground with speed $u$ at angle $\theta$. Let $\vec{L}$ be the angular momentum of the ball about the point of projection at time $t$. Which of the following statements is/are correct?
(A) The torque acting on the ball about the origin is $\vec{\tau} = -mg u\cos\theta \cdot t\hat{k}$.
(B) The magnitude of angular momentum about the origin is $L(t) = \frac{1}{2}mg u\cos\theta \cdot t^2$.
(C) The angular momentum of the projectile about the origin is conserved throughout the flight.
(D) $L(t)$ increases monotonically with time throughout the flight.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A projectile is fired from the ground with initial speed $u = 20\text{ m/s}$ at an angle of $45^\circ$ to the horizontal. Taking $g = 10\text{ m/s}^2$, the horizontal range of the projectile in meters is $R$. Find $R$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A particle is projected from the ground such that its range $R$ and maximum height $H$ satisfy $R = 4\sqrt{3}H$. If the angle of projection with the horizontal is $\theta$ degrees, determine the integer value of $\theta$.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (A) | 3. (C) | 4. (A) | 5. (A) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, D) | 9. 40 | 10. 30
Solution 1:
At the apex, vertical velocity is zero: $v = u\cos(60^\circ) = \frac{u}{2}$.
Kinetic energy: $K = \frac{1}{2}m v^2 = \frac{1}{2}m\left(\frac{u}{2}\right)^2 = \frac{1}{4}\left(\frac{1}{2}mu^2\right) = \frac{K_0}{4}$.
Correct Answer: (B)
Solution 2:
Using $R\tan\theta = 4H \implies \tan\theta = \frac{4H}{R}$.
Given $R = 4\sqrt{3}H \implies \tan\theta = \frac{4H}{4\sqrt{3}H} = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ$.
Correct Answer: (A)
Solution 3:
$\tan\theta = \frac{4H}{R}$. For $R = H$, $\tan\theta = \frac{4H}{H} = 4 \implies \theta = \arctan(4)$.
Correct Answer: (C)
Solution 4:
$H_1 = \frac{u^2\sin^2\theta}{2g}, \quad H_2 = \frac{u^2\cos^2\theta}{2g}$.
$H_1 H_2 = \frac{u^4\sin^2\theta\cos^2\theta}{4g^2} = \frac{(u^2\sin 2\theta / g)^2}{16} = \frac{R^2}{16} \implies R = 4\sqrt{H_1 H_2}$.
Correct Answer: (A)
Solution 5:
At the apex, velocity is horizontal ($v = u\cos\theta$) and acceleration is vertical ($a_n = g$).
$R_c = \frac{v^2}{a_n} = \frac{(u\cos\theta)^2}{g} = \frac{u^2\cos^2\theta}{g}$.
Correct Answer: (A)
Solution 6:
$y = \sqrt{3}x – 2x^2 = x\tan\theta\left(1 – \frac{x}{R}\right) = \sqrt{3}x\left(1 – \frac{2}{\sqrt{3}}x\right) = \sqrt{3}x\left(1 – \frac{x}{\sqrt{3}/2}\right)$.
– $\tan\theta = \sqrt{3} \implies \theta = 60^\circ$ (A is True).
– Range $R = \frac{\sqrt{3}}{2}\text{ m}$ (B is True).
– Maximum height $H = \frac{R\tan\theta}{4} = \frac{(\sqrt{3}/2)(\sqrt{3})}{4} = \frac{3/2}{4} = \frac{3}{8}\text{ m}$ (D is True).
– From trajectory equation coefficient: $\frac{g}{2u^2\cos^2(60^\circ)} = 2 \implies \frac{10}{2u^2(1/4)} = 2 \implies \frac{20}{u^2} = 2 \implies u^2 = 10 \implies u = \sqrt{10}\text{ m/s}$ (C is True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) $\Delta\vec{p} = m(\vec{v}_f – \vec{v}_i) = m[(u\cos\theta\hat{i} – u\sin\theta\hat{j}) – (u\cos\theta\hat{i} + u\sin\theta\hat{j})] = -2mu\sin\theta\hat{j}$ (True).
– (B) $\vec{u}\cdot\vec{v} = 0 \implies t = \frac{u}{g\sin\theta}$ (True).
– (C) $t \le T \implies \frac{u}{g\sin\theta} \le \frac{2u\sin\theta}{g} \implies \sin^2\theta \ge 1/2 \implies \theta \ge 45^\circ$ (True).
– (D) $\vec{v}_{\text{avg}} = \frac{\vec{R}}{T} = \frac{R\hat{i}}{T} = u\cos\theta\hat{i}$ (True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– Force: $\vec{F} = -mg\hat{j}$. Torque about origin: $\vec{\tau} = \vec{r} \times \vec{F} = (x\hat{i} + y\hat{j}) \times (-mg\hat{j}) = -mg x\hat{k} = -mg(u\cos\theta\cdot t)\hat{k}$ (A is True).
– Angular momentum: $\vec{L} = \int \vec{\tau}\,dt = -\frac{1}{2}mg u\cos\theta \cdot t^2\hat{k}$ (B is True).
– Since torque is non-zero, angular momentum is NOT conserved (C is False).
– Magnitude $L \propto t^2$, which increases monotonically (D is True).
Correct Answer: (A, B, D)
Solution 9:
$R = \frac{u^2\sin(2\theta)}{g} = \frac{20^2\sin(90^\circ)}{10} = \frac{400 \times 1}{10} = 40\text{ meters}$.
Correct Answer: 40
Solution 10:
$\tan\theta = \frac{4H}{R} = \frac{4H}{4\sqrt{3}H} = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ$.
Correct Answer: 30