Concept Card: Horizontal Projectile Motion
1. Physical Framework & Coordinate Setup:
A projectile launched horizontally with initial speed $u$ from a height $h$ above level ground moves under the sole influence of gravity (neglecting air resistance).
Choosing the point of projection as the origin $O(0, 0)$, with the horizontal forward direction as $+x$ and the vertically downward direction as $+y$:
- Initial Velocities: $u_x = u, \quad u_y = 0$
- Accelerations: $a_x = 0, \quad a_y = +g$
2. Component Kinematics:
- Horizontal Motion (Uniform Motion):
Because $a_x = 0$, the horizontal velocity remains strictly constant throughout the entire flight:
$v_x(t) = u$
Horizontal displacement at time $t$: $x(t) = u \cdot t \implies t = \frac{x}{u}$ - Vertical Motion (Free Fall from Rest):
Because $u_y = 0$ and $a_y = g$, the vertical motion is identical to free fall from rest:
$v_y(t) = g t$
Vertical displacement at time $t$: $y(t) = \frac{1}{2} g t^2$
Spatial relation: $v_y^2 = 2 g y \implies v_y = \sqrt{2 g y}$
3. Fundamental Governing Formulae:
- Time of Flight ($T$):
The time required to fall the vertical height $h$:
$h = \frac{1}{2} g T^2 \implies T = \sqrt{\frac{2h}{g}}$
Crucial JEE Insight: The time of flight depends solely on vertical height $h$ and $g$; it is completely independent of the projection speed $u$. A body fired horizontally at $1000\text{ m/s}$ and a body dropped simultaneously from the same height hit flat ground at the exact same instant! - Horizontal Range ($R$):
The horizontal distance covered during the flight time $T$:
$R = u \cdot T = u \sqrt{\frac{2h}{g}}$ - Equation of the Trajectory (Path Equation):
Substitute $t = \frac{x}{u}$ into $y = \frac{1}{2} g t^2$:
$y = \frac{1}{2} g \left(\frac{x}{u}\right)^2 = \left(\frac{g}{2 u^2}\right) x^2$
This is of the form $y = k x^2$, proving the trajectory is a parabola opening downwards with vertex at the projection origin. - Instantaneous Velocity & Direction:
$\vec{v}(t) = u\hat{i} + gt\hat{j}$
Speed: $v(t) = \sqrt{u^2 + (gt)^2} = \sqrt{u^2 + 2gy}$
Direction (angle $\beta$ below horizontal):
$\tan\beta = \frac{v_y}{v_x} = \frac{gt}{u} = \frac{\sqrt{2gy}}{u}$ - Velocity on Striking the Ground ($t = T$):
$v_{\text{impact}} = \sqrt{u^2 + 2gh}$
$\tan\beta_{\text{impact}} = \frac{\sqrt{2gh}}{u}$ - Radius of Curvature ($R_c$):
At any point, $R_c = \frac{v^2}{a_n} = \frac{(u^2 + g^2 t^2)^{3/2}}{g u}$.
At the projection point ($t = 0$): $R_{c0} = \frac{u^2}{g}$.
Solved Examples
Example 1 (Flight Time, Range & Landing Velocity):
A ball is kicked horizontally with a speed $u = 98\text{ m/s}$ from the top of a cliff $h = 490\text{ m}$ high. Taking $g = 9.8\text{ m/s}^2$:
(a) Find the time taken by the ball to reach the ground.
(b) Find the horizontal range $R$ from the base of the cliff.
(c) Find the magnitude and direction of velocity upon striking the ground.
Solution:
(a) Time of flight:
$T = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 490}{9.8}} = \sqrt{\frac{980}{9.8}} = \sqrt{100} = 10\text{ seconds}$.
(b) Horizontal range:
$R = u \cdot T = 98\text{ m/s} \times 10\text{ s} = 980\text{ meters}$.
(c) Velocity components at impact:
$v_x = u = 98\text{ m/s}$
$v_y = g T = 9.8 \times 10 = 98\text{ m/s}$
Magnitude: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{98^2 + 98^2} = 98\sqrt{2}\text{ m/s} \approx 138.6\text{ m/s}$.
Direction: $\tan\beta = \frac{v_y}{v_x} = \frac{98}{98} = 1 \implies \beta = 45^\circ$ below the horizontal.
Example 2 (Condition for Velocity Incline & Radius of Curvature):
A particle is projected horizontally with velocity $u = 20\text{ m/s}$ from a high tower. Taking $g = 10\text{ m/s}^2$:
(a) Find the time elapsed when the velocity vector makes an angle of $45^\circ$ with the horizontal.
(b) Find the radius of curvature of the trajectory at $t = 0$ and at the instant when the velocity vector is at $45^\circ$.
Solution:
(a) At any time $t$, the angle $\beta$ with the horizontal satisfies $\tan\beta = \frac{gt}{u}$.
For $\beta = 45^\circ$:
$\tan(45^\circ) = \frac{gt}{u} \implies 1 = \frac{10t}{20} \implies t = 2.0\text{ seconds}$.
(b) Radius of curvature $R_c = \frac{v^2}{a_n}$:
– At $t = 0$: $v = u = 20\text{ m/s}$. Since velocity is horizontal and $\vec{g}$ is vertical, $\vec{g} \perp \vec{v} \implies a_n = g = 10\text{ m/s}^2$.
$R_{c0} = \frac{u^2}{g} = \frac{20^2}{10} = \frac{400}{10} = 40\text{ meters}$.
– At $t = 2\text{ s}$ ($\beta = 45^\circ$):
Speed: $v = \sqrt{u^2 + (gt)^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2}\text{ m/s}$.
Normal acceleration component perpendicular to $\vec{v}$: $a_n = g\cos\beta = 10\cos(45^\circ) = \frac{10}{\sqrt{2}} = 5\sqrt{2}\text{ m/s}^2$.
$R_c = \frac{v^2}{a_n} = \frac{(20\sqrt{2})^2}{5\sqrt{2}} = \frac{800}{5\sqrt{2}} = \frac{160}{\sqrt{2}} = 80\sqrt{2}\text{ m} \approx 113.1\text{ meters}$.
Example 3 (Two Bodies Projected Horizontally in Opposite Directions):
Two stones are projected horizontally in opposite directions from the top of a tower with speeds $u_1 = 30\text{ m/s}$ and $u_2 = 40\text{ m/s}$ at $t = 0$. Taking $g = 10\text{ m/s}^2$:
(a) Find the time $t$ when their velocity vectors become mutually perpendicular ($\vec{v}_1 \perp \vec{v}_2$).
(b) Calculate the horizontal and direct separation between the two stones at this instant.
Solution:
(a) Velocity vectors at time $t$ (taking right as $+\hat{i}$ and down as $+\hat{j}$):
$\vec{v}_1 = u_1\hat{i} + gt\hat{j} = 30\hat{i} + 10t\hat{j}$
$\vec{v}_2 = -u_2\hat{i} + gt\hat{j} = -40\hat{i} + 10t\hat{j}$
For mutual perpendicularity: $\vec{v}_1 \cdot \vec{v}_2 = 0$.
$(30)(-40) + (10t)(10t) = 0 \implies -1200 + 100t^2 = 0 \implies 100t^2 = 1200 \implies t^2 = 12$.
$t = \sqrt{12} = 2\sqrt{3}\text{ seconds} \approx 3.464\text{ s}$.
General Formula: $t = \frac{\sqrt{u_1 u_2}}{g} = \frac{\sqrt{30 \times 40}}{10} = \frac{\sqrt{1200}}{10} = 2\sqrt{3}\text{ s}$.
(b) Horizontal separation:
$x_{\text{rel}} = (u_1 + u_2)t = (30 + 40)(2\sqrt{3}) = 70(2\sqrt{3}) = 140\sqrt{3}\text{ m} \approx 242.49\text{ m}$.
Since both stones fall the same vertical distance ($y_1 = y_2 = \frac{1}{2}gt^2$), their vertical separation is zero ($\Delta y = 0$).
Thus, the direct distance between them is equal to the horizontal separation: $D = 140\sqrt{3}\text{ meters}$.
Example 4 (Aid Package Dropped from a Flying Aircraft):
A relief aircraft flies horizontally at an altitude of $h = 1960\text{ m}$ with a constant speed $u = 720\text{ km/h}$ ($200\text{ m/s}$) heading directly towards a flood survival camp. Taking $g = 9.8\text{ m/s}^2$:
(a) At what horizontal distance from the camp must the cargo be dropped to score a direct hit?
(b) What is the line-of-sight angle (angle below horizontal) from the aircraft to the camp at the instant of release?
(c) Where is the aircraft when the cargo impacts the ground?
Solution:
(a) Time of flight of the dropped package:
$T = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 1960}{9.8}} = \sqrt{\frac{3920}{9.8}} = \sqrt{400} = 20\text{ seconds}$.
Horizontal advance during descent:
$R = u \cdot T = 200\text{ m/s} \times 20\text{ s} = 4000\text{ meters} = 4.0\text{ km}$.
The pilot must release the cargo when the horizontal distance to the camp is $4.0\text{ km}$.
(b) Line-of-sight angle $\theta$ below horizontal:
$\tan\theta = \frac{h}{R} = \frac{1960\text{ m}}{4000\text{ m}} = 0.49 \implies \theta = \arctan(0.49) \approx 26.1^\circ$.
(c) Position of aircraft at impact ($t = 20\text{ s}$):
Since the aircraft moves horizontally at the same constant speed $u = 200\text{ m/s}$, the distance flown by the plane is $u \cdot T = 4000\text{ m}$.
Therefore, the aircraft is directly vertically above the cargo and target camp when the cargo hits the ground!
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A bullet is fired horizontally from a rifle at a target $100\text{ m}$ away, and another bullet is dropped simultaneously from the same height. If air resistance is neglected, which bullet hits the ground first?
(A) The fired bullet hits first
(B) The dropped bullet hits first
(C) Both bullets hit the ground at the exact same instant
(D) Depends on the caliber and mass of the bullets
Problem 2 (JEE Main – Single Correct):
A ball is projected horizontally from the top of a tower with speed $u$. The time when its velocity vector makes an angle of $30^\circ$ with the horizontal is:
(A) $\frac{u}{\sqrt{3}g}$
(B) $\frac{\sqrt{3}u}{g}$
(C) $\frac{u}{g}$
(D) $\frac{2u}{\sqrt{3}g}$
Problem 3 (JEE Main – Single Correct):
An object is thrown horizontally from a height $h$ with initial velocity $u$. The speed with which it strikes the ground is:
(A) $\sqrt{u^2 + 2gh}$
(B) $u + \sqrt{2gh}$
(C) $\sqrt{u^2 + gh}$
(D) $\sqrt{2gh – u^2}$
Problem 4 (JEE Main – Single Correct):
If the projection speed of a horizontal projectile is doubled while keeping the launch height unchanged, its horizontal range will:
(A) Remain unchanged
(B) Double
(C) Quadruple
(D) Increase by $\sqrt{2}$ times
Problem 5 (JEE Main – Single Correct):
A body is projected horizontally from a tower of height $H$ with speed $u$. The radius of curvature of its trajectory at the point of projection is:
(A) $\frac{u^2}{2g}$
(B) $\frac{u^2}{g}$
(C) $\frac{2u^2}{g}$
(D) $\frac{u}{\sqrt{g}}$
Problem 6 (JEE Advanced – One or More Correct):
Two particles $A$ and $B$ are projected horizontally in opposite directions from the top of a high cliff with speeds $u_A$ and $u_B$ respectively. Which of the following statements is/are correct?
(A) Their relative acceleration is zero ($\vec{a}_{\text{rel}} = \vec{0}$).
(B) The relative velocity vector between them remains constant with time.
(C) Their velocity vectors become perpendicular to each other at time $t = \frac{\sqrt{u_A u_B}}{g}$.
(D) Their vertical separation is always zero before either hits the ground.
Problem 7 (JEE Advanced – One or More Correct):
A bomb is released from an airplane flying horizontally with uniform speed $u$ at an altitude $h$. If air resistance is neglected:
(A) The trajectory of the bomb as observed by an observer on the ground is a parabola.
(B) The trajectory of the bomb as observed by the pilot of the airplane is a vertical straight line.
(C) The airplane is directly above the bomb at all times during the fall.
(D) The horizontal range on the ground is $u\sqrt{\frac{2h}{g}}$.
Problem 8 (JEE Advanced – One or More Correct):
For a projectile thrown horizontally with speed $u$ from height $h$:
(A) The tangential acceleration is $a_t = \frac{g^2 t}{\sqrt{u^2 + g^2 t^2}}$.
(B) The normal acceleration is $a_n = \frac{g u}{\sqrt{u^2 + g^2 t^2}}$.
(C) The radius of curvature increases monotonically with time.
(D) The angle made by the velocity with the horizontal continuously approaches $90^\circ$ as time progresses.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A stone is thrown horizontally with a velocity of $15.0\text{ m/s}$ from the top of a $20.0\text{ m}$ high tower. Taking $g = 10\text{ m/s}^2$, the horizontal range of the stone on the ground in meters is $R$. Find $R$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A body is projected horizontally with speed $u = 10\text{ m/s}$ from the top of a tower. After $t\text{ seconds}$, its velocity vector makes an angle of $45^\circ$ with the horizontal. Taking $g = 10\text{ m/s}^2$, determine the numerical value of $10 \times t$.
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (A) | 3. (A) | 4. (B) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 30 | 10. 10
Solution 1:
Vertical motion is governed by $h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}}$. Since both bullets have zero initial vertical velocity ($u_y = 0$) and fall from the same height, their times of flight are identical.
Correct Answer: (C)
Solution 2:
$\tan(30^\circ) = \frac{v_y}{v_x} = \frac{gt}{u} \implies \frac{1}{\sqrt{3}} = \frac{gt}{u} \implies t = \frac{u}{\sqrt{3}g}$.
Correct Answer: (A)
Solution 3:
$v_x = u$, $v_y = \sqrt{2gh}$.
Speed $v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 + 2gh}$.
(Alternatively, by Work-Energy Theorem: $\frac{1}{2}mv^2 – \frac{1}{2}mu^2 = mgh \implies v = \sqrt{u^2 + 2gh}$).
Correct Answer: (A)
Solution 4:
$R = u \sqrt{\frac{2h}{g}}$. Since $h$ is constant, $R \propto u$. Doubling $u$ doubles $R$.
Correct Answer: (B)
Solution 5:
At $t = 0$, velocity is horizontal ($v = u$) and acceleration is vertical ($a = g$).
Since $\vec{v} \perp \vec{g}$, the entire acceleration is normal ($a_n = g$).
$R_{c0} = \frac{v^2}{a_n} = \frac{u^2}{g}$.
Correct Answer: (B)
Solution 6:
– (A) $\vec{a}_{\text{rel}} = \vec{g} – \vec{g} = 0$ (True).
– (B) Since $\vec{a}_{\text{rel}} = 0$, $\vec{v}_{\text{rel}}$ is constant (True).
– (C) $\vec{v}_1 \cdot \vec{v}_2 = (u_A)(-u_B) + (gt)(gt) = -u_A u_B + g^2 t^2 = 0 \implies t = \frac{\sqrt{u_A u_B}}{g}$ (True).
– (D) Both start at same height and fall with $y = \frac{1}{2}gt^2$, so $\Delta y = 0$ (True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) Ground observer sees horizontal velocity $u$ and downward acceleration $g$, tracing parabola $y = \frac{g}{2u^2}x^2$ (True).
– (B) In airplane frame, bomb has zero horizontal relative speed ($v_{\text{rel},x} = u – u = 0$), so it falls along vertical straight line (True).
– (C) Plane and bomb have identical horizontal position $x = ut$ (True).
– (D) Range $R = u T = u\sqrt{\frac{2h}{g}}$ (True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
$\vec{v} = u\hat{i} + gt\hat{j}, \vec{a} = g\hat{j}$.
– $a_t = \frac{\vec{a}\cdot\vec{v}}{v} = \frac{g(gt)}{\sqrt{u^2 + g^2 t^2}} = \frac{g^2 t}{\sqrt{u^2 + g^2 t^2}}$ (A is True).
– $a_n = \sqrt{g^2 – a_t^2} = \frac{gu}{\sqrt{u^2 + g^2 t^2}}$ (B is True).
– $R_c = \frac{v^2}{a_n} = \frac{(u^2 + g^2 t^2)^{3/2}}{gu}$, which increases as $t$ increases (C is True).
– $\tan\beta = \frac{gt}{u} \to \infty \implies \beta \to 90^\circ$ as $t \to \infty$ (D is True).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
Time of flight: $T = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 20}{10}} = \sqrt{4} = 2.0\text{ s}$.
Horizontal range: $R = u \cdot T = 15.0 \times 2.0 = 30\text{ meters}$.
Correct Answer: 30
Solution 10:
$\tan(45^\circ) = \frac{gt}{u} \implies 1 = \frac{10t}{10} \implies t = 1.0\text{ s}$.
Numerical value: $10 \times t = 10 \times 1.0 = 10$.
Correct Answer: 10