Motion in a Plane — JEE Main practice

20 questions

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Sample questions with solutions

Q1 · 2026

A river of width 200m200\mathrm{m} is flowing from west to east with a speed of 18km/h18\mathrm{km/h}. A boat, moving with speed of 36km/h36\mathrm{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.

    20s20\mathrm{s} and 100m100\mathrm{m}

  • B.

    40s40\mathrm{s} and 100m100\mathrm{m}

  • C.

    40s40\mathrm{s} and 200m200\mathrm{m}

  • D.

    40s40\mathrm{s} and 0m0\mathrm{m}

Answer: C

  1. Convert the given speeds into SI units.

    𝑣𝑟=18×518=5m/s,𝑣_𝑟 =18\times\frac{5}{18} =5\mathrm{m/s},

    where 𝑣𝑟𝑣_𝑟 is the speed of the river.

    The speed of the boat in still water is

    𝑣𝑏=36×518=10m/s.𝑣_𝑏 = 36\times\frac{5}{18} =10\mathrm{m/s}.

  2. The width of the river is

    𝑤=200,m.𝑤=200,\mathrm{m}.

    The time required to cross the river depends only on the component of the boat's velocity perpendicular to the banks.

  3. The crossing time is

    𝑡=𝑤𝑣𝑏sin⁡θ,𝑡=\frac{𝑤}{𝑣_𝑏\sin\theta},

    where θ\theta is the angle made by the boat with the bank.

    The time is minimum when the boat moves straight across the river, i.e.,

    θ=90∘.\theta=90^\circ.

  4. Hence, the minimum time for one crossing is

    𝑡1→2=2001020s.𝑡_{1\rightarrow2} = \frac{200}{10} 20\mathrm{s}.

    Similarly, the return trip also takes

    𝑡2→1=20s.𝑡_{2\rightarrow1} = 20\mathrm{s}.

  5. Therefore, the total minimum time is

    𝑇=20+2040s.𝑇 = 20+20 40\mathrm{s}.

  6. During each crossing, the river carries the boat downstream.

    The downstream drift in one crossing is

    Drift=𝑣𝑟×𝑡=5×20=100m.\text{Drift} = 𝑣_𝑟\times 𝑡 = 5\times20 =100\mathrm{m}.

  7. Since the river flows in the same direction during both crossings, the total displacement along the bank is

    100+100=200m.100+100 = 200\mathrm{m}.

  8. Therefore, the minimum time and displacement along the river bank are

    40s and 200m\boxed{40\mathrm{s}\text{ and }200\mathrm{m}}

    Hence, the correct option is C.

Q2 · 2026

A projectile is thrown upward at an angle 60∘60^\circ with the horizontal. The speed of the projectile is 20,m/s20,\mathrm{m/s} when its direction of motion is 45∘45^\circ with the horizontal. The initial speed of the projectile is ‾ m/s\underline{\hspace{2cm}}\,\mathrm{m/s}.

  • A.

    20320\sqrt{3}

  • B.

    40

  • C.

    20220\sqrt{2}

  • D.

    40240\sqrt{2}

Answer: C

  1. A projectile moves with independent horizontal and vertical components of velocity.

    • The horizontal acceleration is zero.
    • Therefore, the horizontal component of velocity remains constant throughout the motion.

  2. Let the initial speed be 𝑢𝑢.

    The initial horizontal component of velocity is

    𝑢𝑥=𝑢cos⁡60∘.𝑢_𝑥=𝑢\cos60^\circ.

  3. At the instant when the projectile is moving at 45∘45^\circ to the horizontal, its speed is 20,m/s20,\mathrm{m/s}.

    Hence, its horizontal component of velocity is

    𝑣𝑥=20cos⁡45∘.𝑣_𝑥=20\cos45^\circ.

  4. Since the horizontal component remains unchanged,

    𝑢cos⁡60∘=20cos⁡45∘.𝑢\cos60^\circ=20\cos45^\circ.

  5. Substitute the trigonometric values:

    𝑢(12)=20(12).𝑢\left(\frac{1}{2}\right) = 20\left(\frac{1}{\sqrt{2}}\right).

  6. Solving for 𝑢𝑢,

    𝑢=20⋅22=202,m/s.𝑢 = 20\cdot\frac{2}{\sqrt{2}} =20\sqrt{2},\mathrm{m/s}.

  7. Therefore,

    𝑢=202m/s\boxed{𝑢=20\sqrt{2}\mathrm{m/s}}

    Hence, the correct option is C.

Q3 · 2026

A boy throws a ball into air at 45∘45^\circ from the horizontal to land it on a roof of a building of height HH. If the ball attains maximum height in 2,s2,\mathrm{s} and lands on the building in 3,s3,\mathrm{s} after launch, then value of HH is ‾ m\underline{\hspace{2cm}}\,\mathrm{m}.

(g=10,m/s2)\left(g=10,\mathrm{m/s^2}\right)
  • A.

    20

  • B.

    25

  • C.

    10

  • D.

    15

Answer: D

  1. Let the initial speed be 𝑢𝑢 and the angle of projection be 45∘45^\circ.

    The vertical component of the initial velocity is

    𝑢𝑦=𝑢sin⁡45∘.𝑢_𝑦=𝑢\sin45^\circ.

  2. The ball reaches its maximum height after 2,s2,\mathrm{s}.

    At the highest point, the vertical velocity becomes zero. Using the first equation of motion,

    𝑣𝑦=𝑢𝑦−g𝑡.𝑣_𝑦=𝑢_𝑦-g𝑡.

    Substituting 𝑣𝑦=0𝑣_𝑦=0, g=10,m/s2g=10,\mathrm{m/s^2} and 𝑡=2,s𝑡=2,\mathrm{s},

    0=𝑢𝑦−10×2,0=𝑢_𝑦-10\times2,

    giving

    𝑢𝑦=20,m/s.𝑢_𝑦=20,\mathrm{m/s}.

  3. The ball lands on the roof after 3,s3,\mathrm{s}.

    The vertical displacement at this instant is the height of the building, HH.

  4. Using the second equation of motion,

    𝑦=𝑢𝑦𝑡−12g𝑡2,𝑦=𝑢_𝑦𝑡-\frac{1}{2}g𝑡^2,

    substitute 𝑢𝑦=20,m/s𝑢_𝑦=20,\mathrm{m/s}, 𝑡=3,s𝑡=3,\mathrm{s} and g=10,m/s2g=10,\mathrm{m/s^2}:

    H=(20)(3)−12(10)(3)2.H=(20)(3)-\frac{1}{2}(10)(3)^2.

  5. Simplifying,

    H=60−45=15,m.H=60-45=15,\mathrm{m}.

  6. Therefore,

    H=15m.\boxed{H=15\mathrm{m}}.

    Hence, the correct option is D.

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