NEET Previous Year Questions Motion in A Plane

Q 1 :
A ball is projected from point A with velocity $20 {ms}^{- 1}$ at an angle $60 °$ to the horizontal direction. At the highest point B of the path (shown in figure), the velocity $v {ms}^{- 1}$ of the ball will be: [2023]

[IMAGE 11]

$20$
$10 \sqrt{3}$
$Zero$
$10$

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(4)

Given, $u = 20 m/s; θ = 60 °; v = ?$

At highest point, the object doesn't have vertical velocity, it has only horizontal velocity and moves in the horizontal direction.

So, $θ' = 0 °$

At $θ = 60 °,$ horizontal velocity $= v \cos θ'$

[IMAGE 12]---------------------------------------------------------

During motion, its horizontal component remains constant

i.e., $u \cos θ = v \cos θ'$

or $20 \cos 60 ° = v \cos 0 °$

$v = 10 m/s$

The velocity of the projectile at the highest point is 10 m/s.

Q 2 :
A bullet is fired from a gun at the speed of $280 {ms}^{- 1}$ in the direction $30 °$ above the horizontal. The maximum height attained by the bullet is $(g = 9.8 {ms}^{- 2}, \sin 30 ° = 0.5)$ [2023]

1000 m
3000 m
2800 m
2000 m

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(1)

Maximum height attained by projectile, is $H = \frac{u^{2} \sin^{2} θ}{2 g}$

Given that, $u = 280 {ms}^{- 1}$ and $θ = 30 °$

$∴ H = \frac{{(280)}^{2} \times \sin^{2} 30 °}{2 \times 9.8} = \frac{{(280)}^{2} \times (0.25)}{2 \times 9.8} = 1000 m$

Q 3 :
A ball is projected with a velocity, $10 {ms}^{- 1}$ at an angle of $60 °$ with the vertical direction. Its speed at the highest point of its trajectory will be: [2022]

$zero$
$5 \sqrt{3} {ms}^{- 1}$
$5 {ms}^{- 1}$
$10 {ms}^{- 1}$

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(2)

Ball is projected with a velocity, $u = 10 m/s$

Angle of projection, $θ = 90 ° - 60 ° = 30 °$

The velocity of a projectile at the highest point will be $v = u \cos θ = 10 \times \cos 30 °$

$= 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3} m/s$

Q 4 :
A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle $' θ'$ to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection $θ$ , is then given by [2021]

$θ = \sin^{- 1} {(\frac{2 g T^{2}}{π^{2} R})}^{1 / 2}$
$θ = \cos^{- 1} {(\frac{g T^{2}}{π^{2} R})}^{1 / 2}$
$θ = \cos^{- 1} {(\frac{π^{2} R}{g T^{2}})}^{1 / 2}$
$θ = \sin^{- 1} {(\frac{π^{2} R}{g T^{2}})}^{1 / 2}$

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(1)

We know, $T = \frac{2 π R}{u} \Rightarrow u = \frac{2 π R}{T}$

Here : $H_{max} = 4 R$

$H_{max} = \frac{u^{2} \sin^{2} θ}{2 g}$ ...(i)

Putting the value of $u$ and $H_{max}$ in equation (i), we get

$4 R = {(\frac{2 π R}{T})}^{2} \times \frac{\sin^{2} θ}{2 g} \Rightarrow θ = \sin^{- 1} {(\frac{2 g T^{2}}{π^{2} R})}^{1 / 2}$

Q 5 :
A projectile is fired from the surface of the earth with a velocity of $5 {ms}^{- 1}$ and angle $θ$ with the horizontal. Another projectile fired from another planet with a velocity of $3 {ms}^{- 1}$ at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in ${ms}^{- 2}$ ) is (Given $g = 9.8 {ms}^{- 2}$ ) [2014]

3.5
5.9
16.3
110.8

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(1)

The equation of trajectory is $y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ}$

where $θ$ is the angle of projection and $u$ is the velocity with which projectile is projected.

For equal trajectories and for same angles of projection,

$\frac{g}{u^{2}} = constant$

As per question, $\frac{9.8}{5^{2}} = \frac{g'}{3^{2}}$

where $g'$ is acceleration due to gravity on the planet.

$g' = \frac{9.8 \times 9}{25} = 3.5 {ms}^{- 2}$

Topic Question Set

Q 1 :
A ball is projected from point A with velocity $20 {ms}^{- 1}$ at an angle $60 °$ to the horizontal direction. At the highest point B of the path (shown in figure), the velocity $v {ms}^{- 1}$ of the ball will be: [2023]

[IMAGE 11]

Q 2 :
A bullet is fired from a gun at the speed of $280 {ms}^{- 1}$ in the direction $30 °$ above the horizontal. The maximum height attained by the bullet is $(g = 9.8 {ms}^{- 2}, \sin 30 ° = 0.5)$ [2023]

Q 3 :
A ball is projected with a velocity, $10 {ms}^{- 1}$ at an angle of $60 °$ with the vertical direction. Its speed at the highest point of its trajectory will be: [2022]

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Topic Question Set

Q 1 : A ball is projected from point A with velocity 20ms-1 at an angle 60° to the horizontal direction. At the highest point B of the path (shown in figure), the velocity vms-1 of the ball will be: [2023] [IMAGE 11]

Q 2 : A bullet is fired from a gun at the speed of 280ms-1 in the direction 30° above the horizontal. The maximum height attained by the bullet is (g=9.8ms-2,sin30°=0.5) [2023]

Q 3 : A ball is projected with a velocity, 10ms-1 at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be: [2022]

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Q 1 :
A ball is projected from point A with velocity $20 {ms}^{- 1}$ at an angle $60 °$ to the horizontal direction. At the highest point B of the path (shown in figure), the velocity $v {ms}^{- 1}$ of the ball will be: [2023]

[IMAGE 11]

Q 2 :
A bullet is fired from a gun at the speed of $280 {ms}^{- 1}$ in the direction $30 °$ above the horizontal. The maximum height attained by the bullet is $(g = 9.8 {ms}^{- 2}, \sin 30 ° = 0.5)$ [2023]

Q 3 :
A ball is projected with a velocity, $10 {ms}^{- 1}$ at an angle of $60 °$ with the vertical direction. Its speed at the highest point of its trajectory will be: [2022]