A river of width 200 m$200 \text{ m}$ is flowing from west to east with a speed of 18 km/h$18 \text{ km/h}$. A boat, moving with speed of 36 km/h$36 \text{ km/h}$ in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are ________ and ________ respectively.
A.20 s and 100 m$20 \text{ s and } 100 \text{ m}$
B.40 s and 0 m$40 \text{ s and } 0 \text{ m}$
C.40 s and 200 m$40 \text{ s and } 200 \text{ m}$
D.40 s and 100 m$40 \text{ s and } 100 \text{ m}$
Solution & Explanation
Related Formula
tmin = dvbr$$t_{\text{min}} = \frac{d}{v_{br}}$$Drift (displacement along bank) = vᵣ × ttotal$$\text{Drift (displacement along bank)} = v_r \times t_{\text{total}}$$
Core Logic
First, convert speeds to m/s.
River speed vR = 18 × (5)/(18) = 5 m/s$v_R = 18 \times \frac{5}{18} = 5 \text{ m/s}$.
Boat speed in still water vBR = 36 × (5)/(18) = 10 m/s$v_{BR} = 36 \times \frac{5}{18} = 10 \text{ m/s}$.
River width d = 200 m$d = 200 \text{ m}$.
River boat vector diagram for Q39 - JEE Main 2026 Evening
For minimum time across the river, the boat must head exactly perpendicular to the river flow.
Minimum time always ignores drift and maxes out the perpendicular velocity vector. Drift simply becomes Vriver × Ttotal$V_{\text{river}} \times T_{\text{total}}$.
A projectile is thrown upward at an angle 60circ$60^{circ}$ with the horizontal. The speed of the projectile is 20 m/s when its direction of motion is 45circ$45^{circ}$ with the horizontal. The initial speed of the projectile is \_\_\_\_ m/s.
A.40√(2)$40\sqrt{2}$
B. 40
C.20√(3)$20\sqrt{3}$
D.20√(2)$20\sqrt{2}$
Solution
Related Formula
uₓ = u θ = v φ$$u_x = u \cos\theta = v \cos\phi$$
Core Logic
Solution diagram for Q32 - JEE Main 2026 Morning
Since horizontal component of velocity remains constant:
An object is projected with kinetic energy K from a point A at an angle 60°$60^{\circ}$ with the horizontal. The ratio of the difference in kinetic energies points B and C to that at point A (see figure), in the absence of air friction is :
Parabolic path of a projectile marked with start A, peak B, and end C.
A.1:2$1:2$
B.2:3$2:3$
C.1:4$1:4$
D.3:4$3:4$
Solution
Related Formula
K = (1)/(2)mu²$$K = \frac{1}{2}mu^{2}$$Kₓ = (1)/(2)m(u θ)² = K ² θ$$K_{x} = \frac{1}{2}m(u \cos \theta)^{2} = K \cos^{2} \theta$$
Core Logic
For a projectile with no air resistance, horizontal velocity remains constant. At the highest point (B), the vertical velocity is zero, so the kinetic energy is entirely due to the horizontal velocity. At point C, it hits the ground with the same speed as projected, restoring full kinetic energy.
Step 1: Find KE at distinct points
(KE)A = K = (1)/(2)mu²$$(\text{KE})_{A} = K = \frac{1}{2}mu^{2}$$
At maximum height (B), velocity is vB = u 60° = (u)/(2)$v_{B} = u \cos 60^{\circ} = \frac{u}{2}$.
Sees: "KE at highest point" → Substitute K ² θ$K \cos^2 \theta$. For θ = 60°$\theta = 60^{\circ}$, 60° = 1/2$\cos 60^{\circ} = 1/2$, so KB = K/4$K_B = K/4$. The difference with ground is 3K/4$3K/4$.
Chapter Mix
Class 11 Physics: Motion in a Plane
Q30jee_main_2026_23_january_eveningMotion under Gravity
A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 m/s²$3 \, \mathrm{m/s}^{2}$ . At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is 5 m/s. The initial height of the aeroplane is ____ m.
(g = 10 m/s ²$^{2}$ )
A.62.5$62.5$
B.92.5$92.5$
C.20$20$
D.82.5$82.5$
Solution
Related Formula
s = ut + (1)/(2)at²$$s = ut + \frac{1}{2}at^2$$
v = u + at$v = u + at$v² = u² + 2as$v^2 = u^2 + 2as$
Core Logic
The motion is divided into three distinct vertical sections:
Motion under Gravity diagram for Q30 - JEE Main 2026 EveningMotion under Gravity diagram for Q30 - JEE Main 2026 Evening
Free fall phase (A → B$A \to B$)
Deceleration phase with parachute (B → C$B \to C$)
Final descent phase to the ground (C → D$C \to D$)
Step 1: Free Fall (A to B)
Initial velocity u = 0$u = 0$, t = 2 s$t = 2 \, \mathrm{s}$, g = 10 m/s²$g = 10 \, \mathrm{m/s^2}$.
Split multi-stage motion problems into clear intervals. The final velocity of interval N becomes the initial velocity of interval N+1. Calculate displacement for each interval independently and sum them up.
A boy thrown a ball into air at 45°$45^{\circ}$ from the horizontal to land it on a roof of a building of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of H is ____ m.
(g = 10 m/s²$g = 10 \text{ m/s}^{2}$)
A.20$20$
B.10$10$
C.25$25$
D.15$15$
Solution
Related Formula
tmaxheight = (uy)/(g)$$t_{\text{max_height}} = \frac{u_y}{g}$$y = uy t - (1)/(2) g t²$$y = u_y t - \frac{1}{2} g t^2$$
Core Logic
Projectile motion onto a building
The time taken to reach the maximum height is given as 2 s$2 \text{ s}$.
This means:
The ball lands on the roof at t = 3 s$t = 3 \text{ s}$. The height H$H$ at this instant is the vertical displacement y$y$.
Using the kinematic equation for vertical motion:
H = uy t - (1)/(2) g t²$$H = u_y t - \frac{1}{2} g t^2$$H = (20)(3) - (1)/(2)(10)(3)²$$H = (20)(3) - \frac{1}{2}(10)(3)^2$$
H = 60 - 5(9)$H = 60 - 5(9)$
H = 60 - 45 = 15 m$$H = 60 - 45 = 15 \text{ m}$$
Pattern Recognition
The launch angle (45°$45^{\circ}$) is distractor data! Maximum height timing strictly dictates initial vertical velocity, which is all you need for the vertical height calculation.
Chapter Mix
Class 11 Physics: Motion in a Plane
More Kinematics Questions — jee_main_2026_21_jan_evening
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.
