A ball rolls off the top of a stairway with horizontal velocity u$u$. The steps are 0.1 ~m$0.1 \mathrm{~m}$ high and 0.1 ~m$0.1 \mathrm{~m}$ wide. The minimum velocity u$u$ with which that ball just hits the step 5 of the stairway will be √(x) ~ms⁻¹$\sqrt{x} \mathrm{~ms}^{-1}$ where x =$x =$ ________ [use g = 10 ~m/s²$g = 10 \mathrm{~m/s}^2$].
Numerical Answer Type:
Enter a numerical valueAnswer: 2 to 2+4 marks
Solution & Explanation
Related Formula
For a projectile fired horizontally from height h$h$ with speed u$u$:
Horizontal Range R = u · t$$\text{Horizontal Range } R = u \cdot t$$Vertical Displacement h = (1)/(2) g t²$$\text{Vertical Displacement } h = \frac{1}{2} g t^2$$
Core Logic
To just clear step 4 and land on the tread of step 5, the flight trajectory must pass beyond the outer corner boundary vertex of the 4th$4^{\text{th}}$ step.
Therefore, net horizontal distance required to cross 4 complete steps is:
Matching the parameter form u = √(x)$u = \sqrt{x}$, we find:
x = 2$x = 2$
Pattern Recognition
To clear the n$n$-th step, the projectile must safely pass the outer point of step (n-1)$(n-1)$. Treat the geometric coordinates of corners as bounding constraints (x = (n-1)w$x = (n-1)w$, y = (n-1)h$y = (n-1)h$) to configure kinematics instantaneously.
Chapter Mix
Class 11 Physics: Motion in a Plane
Trajectory diagram of a ball clearing step corners on a staircase grid for Q52
More Motion in a Plane 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 Plane Questions — jee_main_2024_29_jan_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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