Concept Card: Motion Under Gravity & Vertical Kinematics
1. Governing Principles & Sign Conventions:
Motion under gravity is a classic case of one-dimensional uniformly accelerated motion where acceleration is strictly constant near the surface of Earth:
$\vec{a} = \vec{g} \approx 9.8\text{ m/s}^2 \quad (\text{often taken as } 10\text{ m/s}^2 \text{ in competitive problems})$
Taking the vertically upward direction as positive ($+y$):
$a = -g$ (directed downwards toward Earth’s center at all times, including at the maximum height!).
2. Case 1: Vertical Upward Projection from Ground ($u > 0$):
- Instantaneous Velocity: $v(t) = u – gt$
- Instantaneous Height: $y(t) = ut – \frac{1}{2}gt^2$
- Velocity-Height Relation: $v^2 = u^2 – 2gy$
- Maximum Height Reached ($H_{\text{max}}$):
At the highest point, the particle momentarily stops ($v = 0$):
$0 = u^2 – 2g H_{\text{max}} \implies H_{\text{max}} = \frac{u^2}{2g}$ - Time of Ascent ($t_a$):
$0 = u – gt_a \implies t_a = \frac{u}{g}$ - Time of Descent ($t_d$):
Falling from rest from $H_{\text{max}}$ to ground: $H_{\text{max}} = \frac{1}{2}gt_d^2 \implies t_d = \sqrt{\frac{2H_{\text{max}}}{g}} = \frac{u}{g}$ - Total Time of Flight ($T$):
$T = t_a + t_d = \frac{2u}{g}$ - Landing Velocity: $v_{\text{ground}} = -u \implies |v_{\text{ground}}| = u$ (in the absence of air resistance, landing speed equals launch speed).
- Two-Time Height Symmetry Property:
A particle passes any intermediate height $h < H_{\text{max}}$ twice — once while ascending ($t_1$) and once while descending ($t_2$):
$h = ut – \frac{1}{2}gt^2 \implies \frac{1}{2}gt^2 – ut + h = 0$
– Sum of roots: $t_1 + t_2 = \frac{2u}{g} = T$
– Product of roots: $t_1 t_2 = \frac{2h}{g} \implies h = \frac{1}{2}g t_1 t_2$
3. Case 2: Free Fall from Rest ($u = 0$ at $y = 0$, taking downward as positive $+y$):
- Velocity after time $t$: $v = gt$
- Distance fallen in time $t$: $y = \frac{1}{2}gt^2$
- Velocity after falling distance $y$: $v = \sqrt{2gy}$
- Time to fall total height $H$: $t = \sqrt{\frac{2H}{g}}$
- Galileo’s Odd-Number Law: Displacements in consecutive equal time intervals ($\Delta t$) starting from rest follow the ratio:
$y_1 : y_2 : y_3 : \dots = 1 : 3 : 5 : 7 : \dots$
Displacement in the $n$-th second: $y_n = \frac{g}{2}(2n – 1)$.
4. Case 3: Projection from a Tower of Height $H$:
Taking the launch point at the top of the tower as origin ($y = 0$):
– Thrown upward with speed $u$: $y(t) = ut – \frac{1}{2}gt^2$. At the ground, $y = -H \implies -H = ut – \frac{1}{2}gt^2$.
– Thrown downward with speed $u$: $-H = -ut – \frac{1}{2}gt^2 \implies H = ut + \frac{1}{2}gt^2$.
– Impact speed on the ground: In all cases, $v^2 = u^2 + 2gH \implies v = \sqrt{u^2 + 2gH}$.
Solved Examples
Example 1 (Intermediate Height & Time Symmetry):
A ball thrown vertically upwards passes a window located at height $h = 40\text{ m}$ above the ground twice, with a time interval of $\Delta t = 4.0\text{ s}$ between the two passages. Taking $g = 10\text{ m/s}^2$, determine:
(a) The initial velocity of projection $u$ from the ground.
(b) The maximum height reached by the ball.
(c) The total time of flight $T$.
Solution:
1. Let $t_1$ be the upward passage time and $t_2$ be the downward passage time.
Given: $t_2 – t_1 = 4.0\text{ s}$.
From the quadratic equation for height: $\frac{1}{2}gt^2 – ut + h = 0 \implies 5t^2 – ut + 40 = 0$.
– Sum of roots: $t_1 + t_2 = \frac{u}{5}$.
– Product of roots: $t_1 t_2 = \frac{40}{5} = 8\text{ s}^2$.
2. Relate difference and product of roots:
$(t_2 – t_1)^2 = (t_1 + t_2)^2 – 4t_1 t_2$
$4^2 = \left(\frac{u}{5}\right)^2 – 4(8) \implies 16 = \frac{u^2}{25} – 32 \implies \frac{u^2}{25} = 48 \implies u^2 = 1200$.
$u = \sqrt{1200} = 20\sqrt{3}\text{ m/s} \approx 34.64\text{ m/s}$.
(a) Initial projection velocity $u = 20\sqrt{3}\text{ m/s}$.
(b) Maximum height: $H_{\text{max}} = \frac{u^2}{2g} = \frac{1200}{2 \times 10} = 60\text{ m}$.
(c) Total time of flight: $T = \frac{2u}{g} = \frac{2(20\sqrt{3})}{10} = 4\sqrt{3}\text{ s} \approx 6.93\text{ s}$.
Example 2 (Three-Body Tower Projection Theorem):
From the top of a tower of height $H$:
– Ball 1 is thrown vertically upward with speed $u$ and hits the ground after time $t_1$.
– Ball 2 is thrown vertically downward with the same speed $u$ and hits the ground after time $t_2$.
– Ball 3 is dropped from rest ($u = 0$) and hits the ground after time $t_3$.
Prove that $t_3 = \sqrt{t_1 t_2}$ and determine $H$ and $u$ in terms of $g, t_1, t_2$.
Solution:
1. For Ball 1 (projected upwards):
$-H = u t_1 – \frac{1}{2}g t_1^2 \implies H = -u t_1 + \frac{1}{2}g t_1^2 \quad \text{— (Eq. 1)}$
2. For Ball 2 (projected downwards):
$H = u t_2 + \frac{1}{2}g t_2^2 \quad \text{— (Eq. 2)}$
3. Multiply Eq. 1 by $t_2$ and Eq. 2 by $t_1$ to eliminate $u$:
$H t_2 = -u t_1 t_2 + \frac{1}{2}g t_1^2 t_2$
$H t_1 = +u t_1 t_2 + \frac{1}{2}g t_1 t_2^2$
Adding both equations:
$H(t_1 + t_2) = \frac{1}{2}g t_1 t_2 (t_1 + t_2) \implies H = \frac{1}{2}g t_1 t_2$.
4. For Ball 3 (dropped from rest):
$H = \frac{1}{2}g t_3^2$.
Equating the expressions for $H$:
$\frac{1}{2}g t_3^2 = \frac{1}{2}g t_1 t_2 \implies t_3^2 = t_1 t_2 \implies t_3 = \sqrt{t_1 t_2}$.
5. To find $u$, subtract Eq. 1 from Eq. 2:
$0 = u(t_1 + t_2) – \frac{1}{2}g(t_1^2 – t_2^2) \implies u(t_1 + t_2) = \frac{1}{2}g(t_1 – t_2)(t_1 + t_2) \implies u = \frac{1}{2}g(t_1 – t_2)$.
Example 3 (Crossing of Two Bodies Under Gravity):
A stone $A$ is dropped from the top of a $200\text{ m}$ high cliff at $t = 0$. At the same instant, another stone $B$ is projected vertically upward from the base of the cliff with speed $u = 50\text{ m/s}$. Taking $g = 10\text{ m/s}^2$:
(a) Find when and where the two stones collide/cross each other.
(b) Determine whether stone $B$ is moving upward or downward at the collision instant.
Solution:
(a) Using relative motion along the vertical line:
Acceleration of $A$: $\vec{a}_A = -g\hat{j}$.
Acceleration of $B$: $\vec{a}_B = -g\hat{j}$.
Relative acceleration: $\vec{a}_{\text{rel}} = \vec{a}_A – \vec{a}_B = 0$ (relative motion is uniform!).
Initial relative velocity of approach:
$v_{\text{rel}} = v_A – v_B = 0 – (-50) = 50\text{ m/s}$.
Initial separation: $d = 200\text{ m}$.
Time of meeting: $t = \frac{d}{v_{\text{rel}}} = \frac{200\text{ m}}{50\text{ m/s}} = 4.0\text{ seconds}$.
Position of meeting:
Height from ground using stone $B$:
$y_B = u t – \frac{1}{2}g t^2 = 50(4) – \frac{1}{2}(10)(4^2) = 200 – 80 = 120\text{ m}$ above the base.
(b) Velocity of stone $B$ at $t = 4\text{ s}$:
$v_B = u – gt = 50 – 10(4) = +10\text{ m/s}$.
Since $v_B > 0$, stone $B$ is still moving vertically upwards when they meet.
Example 4 (Leaking Water Tap & Spatial Separation of Drops):
Water drops fall at regular equal intervals of time from a ceiling shower tap located $H = 5.0\text{ m}$ above the floor. When the $1$-st drop strikes the floor, the $3$-rd drop is just beginning to fall from the tap. Taking $g = 10\text{ m/s}^2$, find:
(a) The time interval $\Delta t$ between drops.
(b) The height of the $2$-nd drop above the floor when the $1$-st drop strikes the floor.
Solution:
(a) Time taken by the $1$-st drop to fall $5.0\text{ m}$ from rest:
$H = \frac{1}{2}g t_1^2 \implies 5.0 = \frac{1}{2}(10) t_1^2 \implies 5 t_1^2 = 5 \implies t_1 = 1.0\text{ s}$.
When drop 1 hits the floor, drop 3 is leaving the tap ($t = 0$).
Thus, two equal time intervals have elapsed during the $1.0\text{ s}$ fall of drop 1:
$2 \Delta t = t_1 = 1.0\text{ s} \implies \Delta t = 0.5\text{ seconds}$.
(b) At this instant, the $2$-nd drop has been falling for a time $t_2 = \Delta t = 0.5\text{ s}$.
Distance fallen by drop 2 from the tap:
$y_2 = \frac{1}{2}g t_2^2 = \frac{1}{2}(10)(0.5)^2 = 5 \times 0.25 = 1.25\text{ m}$.
Height of the $2$-nd drop above the floor:
$h_2 = H – y_2 = 5.0\text{ m} – 1.25\text{ m} = 3.75\text{ meters}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A body is projected vertically upwards with velocity $u$. The ratio of the time taken to traverse the first half of its maximum height to the time taken to traverse the second half of its maximum height is:
(A) $\sqrt{2} – 1$
(B) $\frac{1}{\sqrt{2} – 1}$
(C) $\sqrt{2} + 1$
(D) $\frac{\sqrt{2} – 1}{\sqrt{2}}$
Problem 2 (JEE Main – Single Correct):
A stone is dropped from a balloon rising upwards with a constant velocity of $10\text{ m/s}$ when the balloon is at a height of $40\text{ m}$ above the ground. Taking $g = 10\text{ m/s}^2$, the stone hits the ground after:
(A) $2\text{ s}$
(B) $4\text{ s}$
(C) $6\text{ s}$
(D) $8\text{ s}$
Problem 3 (JEE Main – Single Correct):
A particle is thrown vertically upwards with speed $u$. The distance traveled by the particle in the last second of its upward journey is:
(A) $\frac{g}{2}$
(B) $u – \frac{g}{2}$
(C) $g$
(D) $\frac{u}{2}$
Problem 4 (JEE Main – Single Correct):
A ball is dropped into a deep well of depth $h$. If the speed of sound in air is $c$, the splash of water is heard at the top after a total time $T$ equal to:
(A) $\sqrt{\frac{2h}{g}} + \frac{h}{c}$
(B) $\sqrt{\frac{2h}{g}} + \frac{2h}{c}$
(C) $\frac{2h}{g} + \frac{h}{c}$
(D) $\sqrt{\frac{h}{2g}} + \frac{h}{c}$
Problem 5 (JEE Main – Single Correct):
A body freely falling from the rest covers $\frac{7}{16}$ of the total height in the last second of its fall. The total height from which it was dropped is (take $g = 10\text{ m/s}^2$):
(A) $45\text{ m}$
(B) $80\text{ m}$
(C) $100\text{ m}$
(D) $125\text{ m}$
Problem 6 (JEE Advanced – One or More Correct):
A particle is projected vertically upwards with velocity $u$ from the ground. Which of the following statements is/are TRUE?
(A) The acceleration of the particle is $-g$ at all points along its trajectory, including at the apex where $v = 0$.
(B) The average velocity for the entire round trip from launch to ground is zero.
(C) The average speed for the entire round trip is $\frac{u}{2}$.
(D) The height above the ground is related to the two crossing times $t_1$ and $t_2$ by $h = \frac{1}{2}g t_1 t_2$.
Problem 7 (JEE Advanced – One or More Correct):
Two balls $A$ and $B$ are thrown vertically upwards with different initial speeds $u_A$ and $u_B$ ($u_A > u_B$) from the ground. During their flight before either touches the ground:
(A) The relative velocity $(v_A – v_B)$ remains constant with time.
(B) The relative acceleration is zero.
(C) The separation between the two balls increases linearly with time.
(D) The ball with higher speed spends more time in air.
Problem 8 (JEE Advanced – One or More Correct):
A body is released from a height $H$. In the absence of air resistance, which of the following statements is/are correct?
(A) The ratio of distances fallen in the $1$-st, $2$-nd, and $3$-rd seconds is $1 : 3 : 5$.
(B) The ratio of times taken to fall consecutive equal vertical distances $h$ is $1 : (\sqrt{2} – 1) : (\sqrt{3} – \sqrt{2})$.
(C) The speed acquired upon falling half the total height is $\frac{v_{\text{ground}}}{\sqrt{2}}$.
(D) The acceleration of the falling body depends on its mass.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A stone is dropped from the top of an $80\text{ m}$ high building. Taking $g = 10\text{ m/s}^2$, the time taken by the stone to reach the ground is $T\text{ seconds}$. Find $T$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A particle is projected vertically upwards with initial speed $u = 30\text{ m/s}$. Taking $g = 10\text{ m/s}^2$, the total distance (in meters) traversed by the particle in the $4$-th second of its motion is $d$. Find the numerical value of $10 \times d$.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (B) | 3. (A) | 4. (A) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C) | 9. 4 | 10. 50
Solution 1:
Maximum height: $H = \frac{u^2}{2g}$, and total ascent time: $T_a = \frac{u}{g}$.
Time to fall the second half of height from top ($H/2$ to $0$ with $u = 0$):
$\frac{H}{2} = \frac{1}{2}g t_2^2 \implies t_2 = \sqrt{\frac{H}{g}} = \sqrt{\frac{u^2 / (2g)}{g}} = \frac{u}{\sqrt{2}g}$.
Time to cover the first half of height: $t_1 = T_a – t_2 = \frac{u}{g} – \frac{u}{\sqrt{2}g} = \frac{u}{g}\left(1 – \frac{1}{\sqrt{2}}\right) = \frac{u}{g}\left(\frac{\sqrt{2} – 1}{\sqrt{2}}\right)$.
Ratio: $\frac{t_1}{t_2} = \frac{\frac{u}{g}\frac{\sqrt{2}-1}{\sqrt{2}}}{\frac{u}{\sqrt{2}g}} = \sqrt{2} – 1$.
Correct Answer: (A)
Solution 2:
Initial position: $y = 0$. Initial velocity: $u = +10\text{ m/s}$ (inherited from balloon).
Acceleration: $a = -g = -10\text{ m/s}^2$. Ground is at $y = -40\text{ m}$.
$-40 = 10t – \frac{1}{2}(10)t^2 \implies -40 = 10t – 5t^2 \implies 5t^2 – 10t – 40 = 0 \implies t^2 – 2t – 8 = 0$.
$(t – 4)(t + 2) = 0 \implies t = 4\text{ seconds}$.
Correct Answer: (B)
Solution 3:
By time-reversal symmetry, the distance traversed in the last second of ascent equals the distance fallen in the first second of descent from rest ($u = 0$):
$s = \frac{1}{2}g(1)^2 = \frac{g}{2}$.
Notice this is completely independent of the initial speed $u$!
Correct Answer: (A)
Solution 4:
Time for stone to fall distance $h$: $t_1 = \sqrt{\frac{2h}{g}}$.
Time for sound wave to travel distance $h$ upwards at constant speed $c$: $t_2 = \frac{h}{c}$.
Total time: $T = t_1 + t_2 = \sqrt{\frac{2h}{g}} + \frac{h}{c}$.
Correct Answer: (A)
Solution 5:
Let total fall time be $n$ seconds.
Total distance: $H = \frac{1}{2}g n^2$.
Distance in the last ($n$-th) second: $h_n = \frac{g}{2}(2n – 1)$.
Given: $\frac{h_n}{H} = \frac{7}{16} \implies \frac{\frac{g}{2}(2n – 1)}{\frac{1}{2}g n^2} = \frac{7}{16} \implies \frac{2n – 1}{n^2} = \frac{7}{16}$.
$32n – 16 = 7n^2 \implies 7n^2 – 32n + 16 = 0$.
Factoring: $(7n – 4)(n – 4) = 0 \implies n = 4\text{ s}$ (since $n > 1$).
Total height: $H = \frac{1}{2}(10)(4^2) = 5 \times 16 = 80\text{ meters}$.
Correct Answer: (B)
Solution 6:
– (A) True: Gravity acts continuously downwards regardless of instantaneous velocity.
– (B) True: Net displacement is zero, so $v_{\text{avg}} = 0$.
– (C) True: Total distance is $2 H_{\text{max}} = 2\left(\frac{u^2}{2g}\right) = \frac{u^2}{g}$. Total time is $T = \frac{2u}{g}$. Average speed $= \frac{u^2/g}{2u/g} = \frac{u}{2}$.
– (D) True: Fundamental root product identity $h = \frac{1}{2}g t_1 t_2$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) True: $v_{\text{rel}} = (u_A – gt) – (u_B – gt) = u_A – u_B = \text{constant}$.
– (B) True: $a_{\text{rel}} = (-g) – (-g) = 0$.
– (C) True: $\Delta y(t) = v_{\text{rel}} \cdot t = (u_A – u_B)t$.
– (D) True: Time of flight $T = \frac{2u}{g} \propto u$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– (A) True: Galileo’s odd-number rule for free fall from rest.
– (B) True: $t_n = \sqrt{\frac{2(n h)}{g}} – \sqrt{\frac{2(n-1)h}{g}} = \sqrt{\frac{2h}{g}}(\sqrt{n} – \sqrt{n-1})$.
– (C) True: $v^2 = 2g(H/2) = gH = \frac{1}{2}(2gH) = \frac{1}{2}v_{\text{ground}}^2 \implies v = \frac{v_{\text{ground}}}{\sqrt{2}}$.
– (D) False: In vacuum, gravitational acceleration is independent of mass ($a = g$).
Correct Answer: (A, B, C)
Solution 9:
$H = \frac{1}{2}g T^2 \implies 80 = \frac{1}{2}(10) T^2 \implies 5 T^2 = 80 \implies T^2 = 16 \implies T = 4\text{ s}$.
Correct Answer: 4
Solution 10:
Time of ascent to top: $t_a = \frac{u}{g} = \frac{30}{10} = 3.0\text{ seconds}$.
So the particle reaches maximum height at $t = 3\text{ s}$ and starts falling downwards!
The $4$-th second is the time interval from $t = 3\text{ s}$ to $t = 4\text{ s}$.
During this interval, the particle starts from rest at the top ($u = 0$) and falls freely for $1\text{ second}$:
$d = \frac{1}{2}g(1^2) = \frac{1}{2}(10)(1) = 5.0\text{ meters}$.
Numerical value: $10 \times d = 10 \times 5.0 = 50$.
Correct Answer: 50