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PHXI04:MOTION IN A PLANE

361948 Two projectiles \(A\) and \(B\) thrown with speeds in the ratio \(1:\sqrt 2 \) acquired the same heights. If \(A\) is thrown at an angle of \(45^\circ \) with the horizontal, then angle of projection of \(B\) will be

1 \(60^\circ \)

2 \(0^\circ \)

3 \(45^\circ \)

4 \(30^\circ \)

PHXI04:MOTION IN A PLANE

361949 Assertion :
The trajectory of an object moving under the same acceleration due to gravity can be straight line or a parabola depending on the initial conditions.
Reason :
The shape of the trajectory of the motion of an object is determined by the acceleration alone.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI04:MOTION IN A PLANE

361950 The trajectory of projectile, projected from the ground is given by \(y=x-\dfrac{x^{2}}{20}\). Where \(x\) and \(y\) are measured in meter. The maximum height attained by the projectile will be

1 \(10\sqrt 2 \;m\)

2 \(10\;m\)

3 \(200\;m\)

4 \(5\;m\)

JEE - 2023

PHXI04:MOTION IN A PLANE

361951 A small ball is to be thrown, so as to just pass through three equal rings of diameter \(2\,m\) and placed in parallel vertical planes at distance \(8\,m\) apart with their highest points at height \(32\,m\) above the point of projection as shown in the figure. if the angle of projection with horizontal is \(\alpha .\) The value of \(\tan \,\alpha \) is
supporting img

1 3

2 1

3 5

4 2

Explanation:

In figure, points \({A, B}\) and \({C}\) are lying on the trajectory of the path of the ball. The equation of trajectory is
supporting img
\(y=x \tan \alpha-\dfrac{g x^{2}}{2 u^{2} \cos ^{2} \alpha}\)
For point \({A}\),
\(32 - 2 = x\tan \alpha - \frac{{5{x^2}}}{{{u^2}{{\cos }^2}\alpha }}\)
Or \(30 = x\tan \alpha - \frac{5}{{{u^2}}}{x^2}\left( {1 + {{\tan }^2}\alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
For point \({B}\)
\(32 = (x + 8)\tan \alpha - \)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{5}{{{u^2}}}{(x + 8)^2}\left( {1 + {{\tan }^2}\alpha } \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\left( 2 \right)\)
For point \({C}\),
\(32 - 2 = (x + 8)\tan \alpha - \)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{5{{(x + 16)}^2}}}{{{u^2}}}\left( {1 + {{\tan }^2}\alpha } \right)\)
\( \Rightarrow 30 = (x + 16)\tan \alpha - \)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{5{{(x + 16)}^2}\left( {1 + {{\tan }^2}\alpha } \right)}}{{{u^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
Subtracting Eq. (1) from Eq. (2),
\(2 = 8\tan \alpha - \frac{5}{{{u^2}}}(1 + \tan \alpha )\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{x^2} + 16x + 64 - {x^2}} \right)\)
\( \Rightarrow \;\;\;2 = 8\tan \alpha - \frac{5}{{{u^2}}}\left( {1 + {{\tan }^2}\alpha } \right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {16x + 64} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 4 \right)\)
Subtracting Eq. (1) from Eq (3),
\(0 = 16\tan \alpha - \frac{{5\left( {1 + {{\tan }^2}\alpha } \right)}}{{{u^2}}}\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{x^2} + 32x + 256 - {x^2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 5 \right)\)
From Eq. (4) and Eq. (5)
\({\tan ^2}\alpha = \frac{{32 \times 2}}{{64}} = 1\,\)
\(\therefore \,\,\,\,\,\,\tan \alpha = 1\)

PHXI04:MOTION IN A PLANE

361948 Two projectiles \(A\) and \(B\) thrown with speeds in the ratio \(1:\sqrt 2 \) acquired the same heights. If \(A\) is thrown at an angle of \(45^\circ \) with the horizontal, then angle of projection of \(B\) will be

1 \(60^\circ \)

2 \(0^\circ \)

3 \(45^\circ \)

4 \(30^\circ \)

PHXI04:MOTION IN A PLANE

361949 Assertion :
The trajectory of an object moving under the same acceleration due to gravity can be straight line or a parabola depending on the initial conditions.
Reason :
The shape of the trajectory of the motion of an object is determined by the acceleration alone.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI04:MOTION IN A PLANE

361950 The trajectory of projectile, projected from the ground is given by \(y=x-\dfrac{x^{2}}{20}\). Where \(x\) and \(y\) are measured in meter. The maximum height attained by the projectile will be

1 \(10\sqrt 2 \;m\)

2 \(10\;m\)

3 \(200\;m\)

4 \(5\;m\)

JEE - 2023

PHXI04:MOTION IN A PLANE

361951 A small ball is to be thrown, so as to just pass through three equal rings of diameter \(2\,m\) and placed in parallel vertical planes at distance \(8\,m\) apart with their highest points at height \(32\,m\) above the point of projection as shown in the figure. if the angle of projection with horizontal is \(\alpha .\) The value of \(\tan \,\alpha \) is
supporting img

1 3

2 1

3 5

4 2

Explanation:

In figure, points \({A, B}\) and \({C}\) are lying on the trajectory of the path of the ball. The equation of trajectory is
supporting img
\(y=x \tan \alpha-\dfrac{g x^{2}}{2 u^{2} \cos ^{2} \alpha}\)
For point \({A}\),
\(32 - 2 = x\tan \alpha - \frac{{5{x^2}}}{{{u^2}{{\cos }^2}\alpha }}\)
Or \(30 = x\tan \alpha - \frac{5}{{{u^2}}}{x^2}\left( {1 + {{\tan }^2}\alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
For point \({B}\)
\(32 = (x + 8)\tan \alpha - \)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{5}{{{u^2}}}{(x + 8)^2}\left( {1 + {{\tan }^2}\alpha } \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\left( 2 \right)\)
For point \({C}\),
\(32 - 2 = (x + 8)\tan \alpha - \)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{5{{(x + 16)}^2}}}{{{u^2}}}\left( {1 + {{\tan }^2}\alpha } \right)\)
\( \Rightarrow 30 = (x + 16)\tan \alpha - \)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{5{{(x + 16)}^2}\left( {1 + {{\tan }^2}\alpha } \right)}}{{{u^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
Subtracting Eq. (1) from Eq. (2),
\(2 = 8\tan \alpha - \frac{5}{{{u^2}}}(1 + \tan \alpha )\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{x^2} + 16x + 64 - {x^2}} \right)\)
\( \Rightarrow \;\;\;2 = 8\tan \alpha - \frac{5}{{{u^2}}}\left( {1 + {{\tan }^2}\alpha } \right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {16x + 64} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 4 \right)\)
Subtracting Eq. (1) from Eq (3),
\(0 = 16\tan \alpha - \frac{{5\left( {1 + {{\tan }^2}\alpha } \right)}}{{{u^2}}}\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{x^2} + 32x + 256 - {x^2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 5 \right)\)
From Eq. (4) and Eq. (5)
\({\tan ^2}\alpha = \frac{{32 \times 2}}{{64}} = 1\,\)
\(\therefore \,\,\,\,\,\,\tan \alpha = 1\)

PHXI04:MOTION IN A PLANE

361948 Two projectiles \(A\) and \(B\) thrown with speeds in the ratio \(1:\sqrt 2 \) acquired the same heights. If \(A\) is thrown at an angle of \(45^\circ \) with the horizontal, then angle of projection of \(B\) will be

1 \(60^\circ \)

2 \(0^\circ \)

3 \(45^\circ \)

4 \(30^\circ \)

PHXI04:MOTION IN A PLANE

361949 Assertion :
The trajectory of an object moving under the same acceleration due to gravity can be straight line or a parabola depending on the initial conditions.
Reason :
The shape of the trajectory of the motion of an object is determined by the acceleration alone.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI04:MOTION IN A PLANE

361950 The trajectory of projectile, projected from the ground is given by \(y=x-\dfrac{x^{2}}{20}\). Where \(x\) and \(y\) are measured in meter. The maximum height attained by the projectile will be

1 \(10\sqrt 2 \;m\)

2 \(10\;m\)

3 \(200\;m\)

4 \(5\;m\)

JEE - 2023

PHXI04:MOTION IN A PLANE

361951 A small ball is to be thrown, so as to just pass through three equal rings of diameter \(2\,m\) and placed in parallel vertical planes at distance \(8\,m\) apart with their highest points at height \(32\,m\) above the point of projection as shown in the figure. if the angle of projection with horizontal is \(\alpha .\) The value of \(\tan \,\alpha \) is
supporting img

1 3

2 1

3 5

4 2

Explanation:

In figure, points \({A, B}\) and \({C}\) are lying on the trajectory of the path of the ball. The equation of trajectory is
supporting img
\(y=x \tan \alpha-\dfrac{g x^{2}}{2 u^{2} \cos ^{2} \alpha}\)
For point \({A}\),
\(32 - 2 = x\tan \alpha - \frac{{5{x^2}}}{{{u^2}{{\cos }^2}\alpha }}\)
Or \(30 = x\tan \alpha - \frac{5}{{{u^2}}}{x^2}\left( {1 + {{\tan }^2}\alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
For point \({B}\)
\(32 = (x + 8)\tan \alpha - \)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{5}{{{u^2}}}{(x + 8)^2}\left( {1 + {{\tan }^2}\alpha } \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\left( 2 \right)\)
For point \({C}\),
\(32 - 2 = (x + 8)\tan \alpha - \)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{5{{(x + 16)}^2}}}{{{u^2}}}\left( {1 + {{\tan }^2}\alpha } \right)\)
\( \Rightarrow 30 = (x + 16)\tan \alpha - \)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{5{{(x + 16)}^2}\left( {1 + {{\tan }^2}\alpha } \right)}}{{{u^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
Subtracting Eq. (1) from Eq. (2),
\(2 = 8\tan \alpha - \frac{5}{{{u^2}}}(1 + \tan \alpha )\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{x^2} + 16x + 64 - {x^2}} \right)\)
\( \Rightarrow \;\;\;2 = 8\tan \alpha - \frac{5}{{{u^2}}}\left( {1 + {{\tan }^2}\alpha } \right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {16x + 64} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 4 \right)\)
Subtracting Eq. (1) from Eq (3),
\(0 = 16\tan \alpha - \frac{{5\left( {1 + {{\tan }^2}\alpha } \right)}}{{{u^2}}}\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{x^2} + 32x + 256 - {x^2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 5 \right)\)
From Eq. (4) and Eq. (5)
\({\tan ^2}\alpha = \frac{{32 \times 2}}{{64}} = 1\,\)
\(\therefore \,\,\,\,\,\,\tan \alpha = 1\)

PHXI04:MOTION IN A PLANE

361948 Two projectiles \(A\) and \(B\) thrown with speeds in the ratio \(1:\sqrt 2 \) acquired the same heights. If \(A\) is thrown at an angle of \(45^\circ \) with the horizontal, then angle of projection of \(B\) will be

1 \(60^\circ \)

2 \(0^\circ \)

3 \(45^\circ \)

4 \(30^\circ \)

PHXI04:MOTION IN A PLANE

361949 Assertion :
The trajectory of an object moving under the same acceleration due to gravity can be straight line or a parabola depending on the initial conditions.
Reason :
The shape of the trajectory of the motion of an object is determined by the acceleration alone.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI04:MOTION IN A PLANE

361950 The trajectory of projectile, projected from the ground is given by \(y=x-\dfrac{x^{2}}{20}\). Where \(x\) and \(y\) are measured in meter. The maximum height attained by the projectile will be

1 \(10\sqrt 2 \;m\)

2 \(10\;m\)

3 \(200\;m\)

4 \(5\;m\)

JEE - 2023

PHXI04:MOTION IN A PLANE

361951 A small ball is to be thrown, so as to just pass through three equal rings of diameter \(2\,m\) and placed in parallel vertical planes at distance \(8\,m\) apart with their highest points at height \(32\,m\) above the point of projection as shown in the figure. if the angle of projection with horizontal is \(\alpha .\) The value of \(\tan \,\alpha \) is
supporting img

1 3

2 1

3 5

4 2

Explanation:

In figure, points \({A, B}\) and \({C}\) are lying on the trajectory of the path of the ball. The equation of trajectory is
supporting img
\(y=x \tan \alpha-\dfrac{g x^{2}}{2 u^{2} \cos ^{2} \alpha}\)
For point \({A}\),
\(32 - 2 = x\tan \alpha - \frac{{5{x^2}}}{{{u^2}{{\cos }^2}\alpha }}\)
Or \(30 = x\tan \alpha - \frac{5}{{{u^2}}}{x^2}\left( {1 + {{\tan }^2}\alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
For point \({B}\)
\(32 = (x + 8)\tan \alpha - \)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{5}{{{u^2}}}{(x + 8)^2}\left( {1 + {{\tan }^2}\alpha } \right){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\left( 2 \right)\)
For point \({C}\),
\(32 - 2 = (x + 8)\tan \alpha - \)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{5{{(x + 16)}^2}}}{{{u^2}}}\left( {1 + {{\tan }^2}\alpha } \right)\)
\( \Rightarrow 30 = (x + 16)\tan \alpha - \)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{5{{(x + 16)}^2}\left( {1 + {{\tan }^2}\alpha } \right)}}{{{u^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
Subtracting Eq. (1) from Eq. (2),
\(2 = 8\tan \alpha - \frac{5}{{{u^2}}}(1 + \tan \alpha )\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{x^2} + 16x + 64 - {x^2}} \right)\)
\( \Rightarrow \;\;\;2 = 8\tan \alpha - \frac{5}{{{u^2}}}\left( {1 + {{\tan }^2}\alpha } \right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {16x + 64} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 4 \right)\)
Subtracting Eq. (1) from Eq (3),
\(0 = 16\tan \alpha - \frac{{5\left( {1 + {{\tan }^2}\alpha } \right)}}{{{u^2}}}\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{x^2} + 32x + 256 - {x^2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 5 \right)\)
From Eq. (4) and Eq. (5)
\({\tan ^2}\alpha = \frac{{32 \times 2}}{{64}} = 1\,\)
\(\therefore \,\,\,\,\,\,\tan \alpha = 1\)

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