Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is ____ m.
(g = 10 ~m/s²)$(g = 10 \mathrm{~m/s}^{2})$
A.2.5$2.5$
B.4.0$4.0$
C.4.2$4.2$
D.3.8$3.8$
Solution & Explanation
Related Formula
h = ut + (1)/(2)gt²$$h = ut + \frac{1}{2}gt^2$$
Core Logic
Since drops fall at regular intervals, find the total time of fall for the first drop and divide it by the number of intervals to get the time gap between successive drops.
The first drop strikes the floor when the sixth drop begins to fall. There are exactly 5 intervals between the 1st and 6th drop.
Time between each drop Δ t = 1 ~sec5 = 0.2 ~sec$\Delta t = \frac{1 \mathrm{~sec}}{5} = 0.2 \mathrm{~sec}$.
Step 3: Distance of the Fourth Drop
The 4th drop has been falling for 2 intervals (since drops 6, 5, 4 mean the 4th drop was released 2 × 0.2 ~s$2 \times 0.2 \mathrm{~s}$ after the 6th drop started... wait, no. The 4th drop was released before the 5th and 6th).
Time of fall for the 4th$4^{\text{th}}$ drop is:
It was released 2 intervals after the 1st drop, so it has been falling for 1.0 - 2(0.2) = 0.6 ~s$1.0 - 2(0.2) = 0.6 \mathrm{~s}$? Wait, the PDF solution says:
"Time of fall for 4th$4^{\text{th}}$ drop is 1 - 0.6 = 0.4 ~sec$1 - 0.6 = 0.4 \mathrm{~sec}$"
Let's trace: 6th drop (0s fall), 5th drop (0.2s fall), 4th drop (0.4s fall).
Distance fallen by 4th$4^{\text{th}}$ drop:
In 'falling drops' problems, if N drops are in the air, there are (N-1) time intervals. The kth$k^{\text{th}}$ drop from the top has fallen for (k-1)$(k-1)$ intervals.
Keywords:#falling water drops#JEE Main 2026 Morning Q30#Motion in a Straight Line JEE Main 2026#Kinematics Equations JEE Main 2026
More Motion in a Straight Line Previous-Year Questions — Page 5
Q8jee_main_2025_24_jan_eveningVelocity and Acceleration
The position vector of a moving body at any instant of time is given as r = (5t² i - 5t j)m$\vec{\mathrm{r}} = (5\mathrm{t}^2\hat{\mathrm{i}} - 5\mathrm{t}\hat{\mathrm{j}})\mathrm{m}$. The magnitude and direction of velocity at t = 2s$t = 2s$ is,
A.5√(15)$5\sqrt{15}$ m/s, making an angle of ⁻¹4$\tan^{-1}4$ with -ve Y axis
B.5√(15)$5\sqrt{15}$ m/s, making an angle of ⁻¹4$\tan^{-1}4$ with +ve X axis
C.5√(17)$5\sqrt{17}$ m/s, making an angle of ⁻¹4$\tan^{-1}4$ with -ve Y axis
D.5√(17)$5\sqrt{17}$ m/s, making an angle of ⁻¹4$\tan^{-1}4$ with +ve X axis
Solution
Related Formula
v = d rdt$$\vec{v} = \frac{d\vec{r}}{dt}$$
Core Logic
Given position vector:
r = 5t² i - 5t j$$\vec{r} = 5t^2\hat{i} - 5t\hat{j}$$
Differentiating with respect to t$t$:
v = 10t i - 5 j$$\vec{v} = 10t\hat{i} - 5\hat{j}$$
Direction analysis: Velocity vector components angle calculation Q8vₓ = 20$v_x = 20$ along +x$+x$ axis, vy = -5$v_y = -5$ along -y$-y$ axis.
Let θ$\theta$ be the angle made with the negative Y$Y$-axis:
Always read the reference axis carefully in direction questions. Here, the angle is measured from the negative Y$Y$-axis, making θ = vₓ / |vy|$\tan\theta = v_x / |v_y|$.
Chapter Mix
Class 11 Physics: Motion in a Plane
Q9jee_main_2025_28_jan_eveningVelocity Time Graphs
The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between t = 0$t = 0$ to t = 4s$t = 4s$ ?
The velocity-time profile tracking motion across consecutive geometric shapes up to 4 seconds.
A.30m$30\mathrm{m}$
B.10m$10\mathrm{m}$
C.13m$13\mathrm{m}$
D.11m$11\mathrm{m}$
Solution
Related Formula
The distance traveled by an object equals the total area enclosed under its velocity-time (v-t$v-t$) plot along the time axis, treating all regional boundaries as strictly positive metrics:
From the geometric grid profile between t = 0$t = 0$ and t = 4 s$t = 4\text{ s}$:
First Region (t=0$t=0$ to t=2 s$t=2\text{ s}$): Forms a \right-angled \triangle with base = 2 s$= 2\text{ s}$ and peak height = 10 ms⁻¹$= 10\text{ ms}^{-1}$.
Second Region (t=2$t=2$ to t=4 s$t=4\text{ s}$): Forms a standard rectangle with width = (4 - 2) = 2 s$= (4 - 2) = 2\text{ s}$ and height = 10 ms⁻¹$= 10\text{ ms}^{-1}$.
Always differentiate between distance and displacement on graph tracks. Displacement treats components below the axis as negative fields, while distance calculates absolute geometric magnitudes without direction.
Chapter Mix
Class 11 Physics: Motion in a Straight Line
Qjee_main_2025_29_jan_morningProjectile Motion
Two projectiles are fired with same initial speed from same point on ground at angles of (45° - α)$(45^{\circ} - \alpha)$ and (45° + α)$(45^{\circ} + \alpha)$ , respectively, with the horizontal direction. The ratio of their maximum heights attained is :
For complementary angles shifted symmetric to 45°$45^{\circ}$, ratios simplify cleanly via 2α$\sin 2\alpha$ identities.
Chapter Mix
Class 11 Physics: Motion in a Plane
Qjee_main_2025_29_jan_morningRelative Motion
The maximum speed of a boat in still water is 27 km/h$27 \, \text{km/h}$ . Now this boat is moving downstream in a river flowing at 9 km/h$9 \, \text{km/h}$ . A man in the boat throws a ball vertically upwards with speed of 10 m/s$10 \, \text{m/s}$ . Range of the ball as observed by an observer at rest on the river bank, is ________ cm. (Take g = 10 m/s²$g = 10 \, \text{m/s}^2$ )
Numerical Answer.Answer: 2000 to 2000
Solution
Related Formula
T = (2uy)/(g), R = vₓ · T$$T = \frac{2u_y}{g}, \quad R = v_x \cdot T$$
Core Logic
Relative range tracking frame vector
The total horizontal velocity components combine due to downstream motion addition
A particle moving in a circle of radius R$R$ with uniform speed takes time T$T$ to complete one revolution. If this particle is projected with the same speed at an angle θ$\theta$ to the horizontal, the maximum height attained by it is equal to 4R$4R$. The angle of projection θ$\theta$ is then given by:
Connect circular metrics directly to projectile parameters via velocity matching. Keeping terms unsimplified makes it easy to cancel common elements later.
Chapter Mix
Class 11 Physics: Motion in a Plane
More Motion in a Straight Line Questions — jee_main_2026_28_january_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
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