JEE Advanced Physics Projectile Motion PYQ | Previous Year Questions with Solutions

Q 1 :

A projectile is given an initial velocity of $(\hat{i} + 2 \hat{j})$ m/s, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g = 10 {m/s}^{2}$ , the equation of its trajectory is: [2012]

$y = x - 5 x^{2}$
$y = 2 x - 5 x^{2}$
$4 y = 2 x - 5 x^{2}$
$4 y = 2 x - 25 x^{2}$

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(2)

$From equation, \vec{v} = \hat{i} + 2 \hat{j}$

$\Rightarrow x = t$ $. . . (i)$

$y = 2 t - \frac{1}{2} (10 t^{2})$ $. . . (i i)$

$From (i) and (ii), y = 2 x - 5 x^{2}$

Q 2 :

A ball is thrown from ground at angle $θ$ with horizontal and with an initial speed $u_{0}$ . For the resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is $V_{1}$ . After hitting the ground, the ball rebounds at the same angle $θ$ but with a reduced speed of $\frac{u_{0}}{a}$ . Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for entire duration of motion is 0.8 $V_{1}$ , the value of $α$ is _____ [2019]

[IMAGE 22]

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(4)

Let $u_{1}, u_{2}, u_{3}, \dots$ be the horizontal velocity of the projectiles and $t_{1}, t_{2}, t_{3}, \dots$ be the time taken as shown in figure.

[IMAGE 23]

$Average velocity = \frac{total displacement}{total time taken}$

$Time of flight T = \frac{2 u \sin θ}{g}$

$For given value of θ, value of T will change with the value of u .$

$∴$ $Total time taken = t_{1} + t_{2} + t_{3} + \dots$

$= t_{1} + \frac{t_{1}}{α} + \frac{t_{1}}{α^{2}} + \dots = \frac{t_{1}}{1 - \frac{1}{α}} = \frac{t_{1} α}{α - 1}$

$Total displacement = u_{1} t_{1} + u_{2} t_{2} + u_{3} t_{3} + \dots$

$= u_{1} t_{1} + \frac{u_{1}}{α} \cdot \frac{t_{1}}{α} + \frac{u_{1}}{α^{2}} \cdot \frac{t_{1}}{α^{2}} + \dots$ $= \frac{u_{1} t_{1}}{1 - \frac{1}{α^{2}}} = \frac{u_{1} t_{1} α^{2}}{α^{2} - 1}$

$∴$ $Average velocity = \frac{total displacement}{total time}$

$= \frac{u_{1} t_{1} α^{2}}{α^{2} - 1} \times \frac{α - 1}{t_{1} α} = \frac{u_{1} α}{α + 1}$

$According to question, average velocity = 0.8 V_{1}$

$\Rightarrow 0.8 V_{1} = \frac{u_{1} α}{α + 1} . . . (i)$

$V_{1} = \frac{u_{1} t_{1}}{t_{1}} = u_{1} . . . (i i)$

$From (i) and (ii):$

$α = 0.8 α + 0.8 or α - 0.8 α = 0.8 or 0.2 α = 0.8$

$or α = \frac{0.8}{0.2} = 4$

Q 3 :

A train is moving along a straight line with a constant acceleration 'a'. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of $60 °$ to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in ${m/s}^{2}$ , is ____. [2011]

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(5)

$Here, u = 10 m/s, θ = 60 °$

[IMAGE 24]

$∴$ $u_{x} = u \cos θ = 10 \times \cos 60 ° = 5 m/s$

$and u_{y} = u \sin θ = 10 \times \sin 60 ° = 5 \sqrt{3} m/s$

$∴$ $t = \frac{2 u_{y}}{g} = \frac{2 \times 5 \sqrt{3}}{10} = \sqrt{3} s$

$Let a be the acceleration of train.$

$1.15 = u_{x} t - \frac{1}{2} a t^{2}$

$\Rightarrow 1.15 = 5 \times \sqrt{3} - \frac{1}{2} \times a \times {(\sqrt{3})}^{2}$

$\Rightarrow \frac{3 a}{2} = 5 \sqrt{3} - 1.15 = 8.65 - 1.15 = 7.5$

$∴$ $a = 7.5 \times \frac{2}{3} = 5 {m/s}^{2}$

Q 4 :

A projectile of mass 200 g is launched in a viscous medium at an angle $60 °$ with the horizontal, with an initial velocity of 270 m/s. It experiences a viscous drag force $\vec{F} = - c \vec{v}$ where the drag coefficient c = 0.1 kg/s and $\vec{v}$ is the instantaneous velocity of the projectile. The projectile hits a vertical wall after 2 s. Taking e = 2.7, the horizontal distance of the wall from the point of projection (in m) is ______. [2025]

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(170)

$Net force$

${\vec{F}}_{net} = m \frac{d \vec{v}}{d t}$

$m \vec{g} + \vec{F} = m \frac{d \vec{v}}{d t}$

[IMAGE 25]

$m \vec{g} - c \vec{v} = m \frac{d \vec{v}}{d t}$ $[∵ {\vec{F}}_{drag} = - c \vec{v} given]$

$Horizontal direction$

$- c v_{x} = m \frac{d v_{x}}{d t}$

$- \frac{c}{m} \int_{0}^{t} d t = \int_{v_{0 x}}^{v_{x}} \frac{d v_{x}}{v_{x}} \Rightarrow - \frac{t}{2} = \ln \frac{v_{x}}{v_{0 x}}$

$\frac{d x}{d t} = v_{x} = v_{0 x} e^{- t / 2}$

$\int_{0}^{S_{x}} d x = v_{0 x} \int_{0}^{t} e^{- t / 2} d t$

$S_{x} = 2 v_{0 x} (1 - e^{- t / 2})$

$At t = 2 s$

$S_{x} = 2 \times 270 \times \cos 60 ° [1 - \frac{1}{e}]$

$S_{x} = 270 (1 - \frac{1}{2.7}) = \frac{270}{2.7} \times 1.7 = 170 m$

Q 5 :

A projectile is fired from horizontal ground with speed $v$ and projection angle $θ$ . When the acceleration due to gravity is $g$ , the range of the projectile is $d$ . If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is $g' = \frac{g}{0.81}$ , then the new range is $d' = n d$ . The value of $n$ is ____ . [2022]

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(0.95)

After entering in new region, time taken by projectile to reach ground is given as

$t = \sqrt{\frac{2 h}{g_{eff}}} = \sqrt{\frac{2 \times 0.81 \times u^{2} \sin^{2} θ}{g \times 2 g}} = 0.9 \frac{u \sin θ}{g}$

So, horizontal displacement done by projectile in new region is given as

$x = 0.9 \times \frac{u \sin θ}{g} \times u \cos θ = 0.9 (\frac{d}{2})$

$Now, new range = \frac{d}{2} + 0.9 \frac{d}{2} = 0.95 d$

Q 6 :

A projectile is thrown from a point O on the ground at an angle $45 °$ from the vertical and with a speed $5 \sqrt{2} m/s.$ The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, $t$ seconds after the splitting, falls to the ground at a distance $x$ meters from the point O. The acceleration due to gravity $g = 10 {m/s}^{2}$ . The value of $t$ is ______. [2021]

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(0.50)

$Range R = \frac{2 u_{x} u_{y}}{g} = \frac{2 \times 5 \times 5}{10} = 5 m$

[IMAGE 26]

$Time of flight T = \frac{2 u_{y}}{g} = \frac{2 \times 5}{10} = 1 s$

[IMAGE 27]

[IMAGE 28]

$∵$ $Time of motion of one part falling vertically downwards = 0.5 s = \frac{T}{2}$

$From law of conservation of momentum, P_{i} = P_{f}$

$2 m \times 5 = m \times v \Rightarrow v = 10 m/s$

As there is no initial velocity in vertical direction for second particle, so it will also take same time as first one.

$∴$ $Time of motion of another part, t = \frac{T}{2} = 0.5 s$

Q 7 :

A projectile is thrown from a point O on the ground at an angle $45 °$ from the vertical and with a speed $5 \sqrt{2} m/s.$ The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, $t$ seconds after the splitting, falls to the ground at a distance $x$ meters from the point O. The acceleration due to gravity $g = 10 {m/s}^{2}$ . The value of $x$ is _________ . [2021]

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(7.5)

$Range R = \frac{2 u_{x} u_{y}}{g} = \frac{2 \times 5 \times 5}{10} = 5 m$

[IMAGE 29]

$Time of flight T = \frac{2 u_{y}}{g} = \frac{2 \times 5}{10} = 1 s$

[IMAGE 30]

[IMAGE 31]

$∵$ $Time of motion of one part falling vertically downwards = 0.5 s = \frac{T}{2}$

$From law of conservation of momentum, P_{i} = P_{f}$

$2 m \times 5 = m \times v \Rightarrow v = 10 m/s$

As there is no initial velocity in vertical direction for second particle, so it will also take same time as first one.

$∴$ $Time of motion of another part, t = \frac{T}{2} = 0.5 s$

$Displacement of other part in 0.5 s in horizontal direction = v \frac{T}{2} = 10 \times 0.5 = 5 m = R$

$∴$ $Total distance of second part from point ‘O’ is,$

$x = \frac{3 R}{2} = 3 \times \frac{5}{2} \Rightarrow x = 7.5 m$

Topic Question Set

Q 1 :

A projectile is given an initial velocity of $(\hat{i} + 2 \hat{j})$ m/s, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g = 10 {m/s}^{2}$ , the equation of its trajectory is: [2012]

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Topic Question Set

Q 1 : A projectile is given an initial velocity of (i^+2j^)m/s, where i^ is along the ground and j^ is along the vertical. If g=10 m/s2, the equation of its trajectory is: [2012]

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Want Us to Email you About Special Offers & Updates?

Q 1 :

A projectile is given an initial velocity of $(\hat{i} + 2 \hat{j})$ m/s, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If $g = 10 {m/s}^{2}$ , the equation of its trajectory is: [2012]