JEE Main Motion in Plane PYQ

11. (JEE Main 2023 (Online) 6th April Morning Shift )

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R

Assertion A : When a body is projected at an angle $45^{\circ}$ , it's range is maximum.

Reason R : For maximum range, the value of $\sin 2 θ$ should be equal to one.

In the light of the above statements, choose the correct answer from the options given below:

A. Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

B. $A$ is true but $R$ is false

C. $A$ is false but $R$ is true

D. Both $A$ and $R$ are correct but $R$ is NOT the correct explanation of $A$

rrect Option is (A)

Assertion A: When a body is projected at an angle of $45^{\circ}$ , its range is maximum. This is true, and it's a well-established fact in physics. The maximum range of a projectile, assuming no air resistance and flat terrain, is achieved at an angle of $45^{\circ}$ .

Reason R: For maximum range, the value of $\sin 2 θ$ should be equal to one. This is also true. The range of a projectile, again assuming no air resistance and flat terrain, can be calculated using the formula $R = (v^{2} / g) \cdot \sin (2 θ)$ , where $v$ is the initial velocity of the projectile, $g$ is the acceleration due to gravity, and $θ$ is the launch angle. For the range to be maximized, $\sin (2 θ)$ must be maximized, and the maximum value of $\sin (2 θ)$ is 1. This occurs when $2 θ = 90$ degrees, or $θ = 45$ degrees, which corresponds to the assertion.

12. (JEE Main 2023 (Online) 6th April Morning Shift )

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R

Assertion A : When a body is projected at an angle $45^{\circ}$ , it's range is maximum.

Reason R : For maximum range, the value of $\sin 2 θ$ should be equal to one.

In the light of the above statements, choose the correct answer from the options given below:

A. Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

B. $A$ is true but $R$ is false

C. $A$ is false but $R$ is true

D. Both $A$ and $R$ are correct but $R$ is NOT the correct explanation of $A$

rrect Option is (A)

Assertion A: When a body is projected at an angle of $45^{\circ}$ , its range is maximum. This is true, and it's a well-established fact in physics. The maximum range of a projectile, assuming no air resistance and flat terrain, is achieved at an angle of $45^{\circ}$ .

Reason R: For maximum range, the value of $\sin 2 θ$ should be equal to one. This is also true. The range of a projectile, again assuming no air resistance and flat terrain, can be calculated using the formula $R = (v^{2} / g) \cdot \sin (2 θ)$ , where $v$ is the initial velocity of the projectile, $g$ is the acceleration due to gravity, and $θ$ is the launch angle. For the range to be maximized, $\sin (2 θ)$ must be maximized, and the maximum value of $\sin (2 θ)$ is 1. This occurs when $2 θ = 90$ degrees, or $θ = 45$ degrees, which corresponds to the assertion.

13. (JEE Main 2023 (Online) 1st February Morning Shift )

A child stands on the edge of the cliff $10 m$ above the ground and throws a stone horizontally with an initial speed of $5 {ms}^{- 1}$ . Neglecting the air resistance, the speed with which the stone hits the ground will be ${ms}^{- 1}$ (given, $g = 10 {ms}^{- 2}$ ).

A. 20

B. 25

C. 30

D. 15

rrect Option is (D)

JEE Main 2023 (Online) 1st February Morning Shift Physics - Motion Question 31 English Explanation

$\begin{aligned} v_{y} = \sqrt{2 g h} = \sqrt{200} \\ v_{n e t} = \sqrt{25 + 200} = 15 m / s \end{aligned}$

14. (JEE Main 2023 (Online) 31st January Morning Shift )

The initial speed of a projectile fired from ground is $u$ . At the highest point during its motion, the speed of projectile is $\frac{\sqrt{3}}{2} u$ . The time of flight of the projectile is :

A. $\frac{u}{g}$

B. $\frac{2 u}{g}$

C. $\frac{u}{2 g}$

D.

rrect Option is (A)

$u \cos θ = \frac{\sqrt{3}}{2} u$

$\Rightarrow$ $\cos θ = \frac{\sqrt{3}}{2}$

$\Rightarrow$ $θ = 30^{\circ}$

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

15. (JEE Main 2023 (Online) 25th January Evening Shift )

Two objects are projected with same velocity 'u' however at different angles $α$ and $β$ with the horizontal. If $α + β = 90^{\circ}$ , the ratio of horizontal range of the first object to the 2nd object will be :

A. 1 : 1

B. 2 : 1

C. 1 : 2

D. 4 : 1

rrect Option is (A)

$Range = \frac{u^{2} \sin 2 θ}{g}$

Range for projection angle " $α$ "

$R_{1} = \frac{u^{2} \sin 2 α}{g}$

Range for projection angle " $β$ "

$\begin{aligned} R_{2} = \frac{u^{2} \sin 2 β}{g} \\ α + β = 90^{\circ} (Given) \\ \Rightarrow β = 90^{\circ} - α \\ R_{2} = \frac{u^{2} \sin 2 (90^{\circ} - α)}{g} \\ R_{2} = \frac{u^{2} \sin (180^{\circ} - 2 α)}{g} \\ R_{2} = \frac{u^{2} \sin 2 α}{g} \\ \Rightarrow \frac{R_{1}}{R_{2}} = \frac{(\frac{u^{2} \sin 2 α}{g})}{(\frac{u^{2} \sin 2 α}{g})} = \frac{1}{1} \end{aligned}$