362017
Assertion : When \(\theta=45^{\circ}\) or \(135^{\circ}\), the value of \(R\) remains the same, only the sign changes. Reason : \(R = \frac{{{u^2}\sin 2\theta }}{g}\)
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.
Explanation:
Horizontal range, \(R=\dfrac{u^{2} \sin 2 \theta}{g}\) When \(\theta=45^{\circ}\), maximum horizontal range, \(R_{\max }=\dfrac{u^{2}}{g} \sin 90^{\circ}=\dfrac{u^{2}}{g}\) When \(\theta=135^{\circ}\), maximum horizontal range, \(R=\dfrac{u^{2}}{g} \sin 270^{\circ}=\dfrac{-u^{2}}{g}\) (Negative sign implies opposite direction)
AIIMS - 2017
PHXI04:MOTION IN A PLANE
362018
For an object thrown at \(45^\circ \) to horizontal, the maximum height (\(H\)) and horizontal range (\(R\)) are related as
362019
Statement A : When the range of projectile is maximum, the time of flight is the largest. Statement B : Range is maximum when angle of projection is \(45^\circ \)
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
\(T = \frac{{2u\sin \theta }}{g}\), it will maximum, when \(\theta = 0^\circ \). \({R_{\max }} = \frac{{{u^2}}}{g},{\rm{for}}\,\,\,\theta = 45^\circ \) So option (2) is correct.
PHXI04:MOTION IN A PLANE
362020
The ceiling of a long hall is 25\(m\) high. Then, the maximum horizontal distance that a ball thrown with a speed of 40 \(m\)/\(s\) can go without hitting the ceiling of the hall, is
1 \(95.5\,m\)
2 \(105.5\,m\)
3 \(100\,m\)
4 \(150.5\,\,m\)
Explanation:
Given, initial velocity \((u) = 40\,m/s\) Height of the hall \((H) = 25\,m\) Let the angle of projection of the ball be \(\theta \), when maximum height attained by it be 25\(m\). Maximum height attained by the ball \(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\) \(25 = \frac{{{{(40)}^2}{{\sin }^2}\theta }}{{2 \times 9.8}}\) or \({\sin ^2}\theta = \frac{{25 \times 2 \times 9.8}}{{1600}}\) \( = 0.3063\) or \(\sin \theta = 0.5534\) \( = \sin 33.6^\circ \) or \(\theta = 33.6^\circ \) \(\therefore \) Horizontal range \((R) = \frac{{{u^2}\sin 2\theta }}{g}\) \( = \frac{{{{(40)}^2}\sin 2 \times 33.6^\circ }}{{9.8}}\) \( = \frac{{1600 \times \sin 67.2^\circ }}{{9.8}}\) \( = \frac{{1600 \times 0.9219}}{{9.8}} = 150.5\,m\)
NCERT Exemplar
PHXI04:MOTION IN A PLANE
362021
The range of the projectile projected at an angle of \(15^{\circ}\) with horizontal is \(50\;m\). If the projectile is projected with same velocity at an angle of \(45^{\circ}\) with horizontal, then its range will be
362017
Assertion : When \(\theta=45^{\circ}\) or \(135^{\circ}\), the value of \(R\) remains the same, only the sign changes. Reason : \(R = \frac{{{u^2}\sin 2\theta }}{g}\)
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.
Explanation:
Horizontal range, \(R=\dfrac{u^{2} \sin 2 \theta}{g}\) When \(\theta=45^{\circ}\), maximum horizontal range, \(R_{\max }=\dfrac{u^{2}}{g} \sin 90^{\circ}=\dfrac{u^{2}}{g}\) When \(\theta=135^{\circ}\), maximum horizontal range, \(R=\dfrac{u^{2}}{g} \sin 270^{\circ}=\dfrac{-u^{2}}{g}\) (Negative sign implies opposite direction)
AIIMS - 2017
PHXI04:MOTION IN A PLANE
362018
For an object thrown at \(45^\circ \) to horizontal, the maximum height (\(H\)) and horizontal range (\(R\)) are related as
362019
Statement A : When the range of projectile is maximum, the time of flight is the largest. Statement B : Range is maximum when angle of projection is \(45^\circ \)
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
\(T = \frac{{2u\sin \theta }}{g}\), it will maximum, when \(\theta = 0^\circ \). \({R_{\max }} = \frac{{{u^2}}}{g},{\rm{for}}\,\,\,\theta = 45^\circ \) So option (2) is correct.
PHXI04:MOTION IN A PLANE
362020
The ceiling of a long hall is 25\(m\) high. Then, the maximum horizontal distance that a ball thrown with a speed of 40 \(m\)/\(s\) can go without hitting the ceiling of the hall, is
1 \(95.5\,m\)
2 \(105.5\,m\)
3 \(100\,m\)
4 \(150.5\,\,m\)
Explanation:
Given, initial velocity \((u) = 40\,m/s\) Height of the hall \((H) = 25\,m\) Let the angle of projection of the ball be \(\theta \), when maximum height attained by it be 25\(m\). Maximum height attained by the ball \(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\) \(25 = \frac{{{{(40)}^2}{{\sin }^2}\theta }}{{2 \times 9.8}}\) or \({\sin ^2}\theta = \frac{{25 \times 2 \times 9.8}}{{1600}}\) \( = 0.3063\) or \(\sin \theta = 0.5534\) \( = \sin 33.6^\circ \) or \(\theta = 33.6^\circ \) \(\therefore \) Horizontal range \((R) = \frac{{{u^2}\sin 2\theta }}{g}\) \( = \frac{{{{(40)}^2}\sin 2 \times 33.6^\circ }}{{9.8}}\) \( = \frac{{1600 \times \sin 67.2^\circ }}{{9.8}}\) \( = \frac{{1600 \times 0.9219}}{{9.8}} = 150.5\,m\)
NCERT Exemplar
PHXI04:MOTION IN A PLANE
362021
The range of the projectile projected at an angle of \(15^{\circ}\) with horizontal is \(50\;m\). If the projectile is projected with same velocity at an angle of \(45^{\circ}\) with horizontal, then its range will be
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI04:MOTION IN A PLANE
362017
Assertion : When \(\theta=45^{\circ}\) or \(135^{\circ}\), the value of \(R\) remains the same, only the sign changes. Reason : \(R = \frac{{{u^2}\sin 2\theta }}{g}\)
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.
Explanation:
Horizontal range, \(R=\dfrac{u^{2} \sin 2 \theta}{g}\) When \(\theta=45^{\circ}\), maximum horizontal range, \(R_{\max }=\dfrac{u^{2}}{g} \sin 90^{\circ}=\dfrac{u^{2}}{g}\) When \(\theta=135^{\circ}\), maximum horizontal range, \(R=\dfrac{u^{2}}{g} \sin 270^{\circ}=\dfrac{-u^{2}}{g}\) (Negative sign implies opposite direction)
AIIMS - 2017
PHXI04:MOTION IN A PLANE
362018
For an object thrown at \(45^\circ \) to horizontal, the maximum height (\(H\)) and horizontal range (\(R\)) are related as
362019
Statement A : When the range of projectile is maximum, the time of flight is the largest. Statement B : Range is maximum when angle of projection is \(45^\circ \)
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
\(T = \frac{{2u\sin \theta }}{g}\), it will maximum, when \(\theta = 0^\circ \). \({R_{\max }} = \frac{{{u^2}}}{g},{\rm{for}}\,\,\,\theta = 45^\circ \) So option (2) is correct.
PHXI04:MOTION IN A PLANE
362020
The ceiling of a long hall is 25\(m\) high. Then, the maximum horizontal distance that a ball thrown with a speed of 40 \(m\)/\(s\) can go without hitting the ceiling of the hall, is
1 \(95.5\,m\)
2 \(105.5\,m\)
3 \(100\,m\)
4 \(150.5\,\,m\)
Explanation:
Given, initial velocity \((u) = 40\,m/s\) Height of the hall \((H) = 25\,m\) Let the angle of projection of the ball be \(\theta \), when maximum height attained by it be 25\(m\). Maximum height attained by the ball \(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\) \(25 = \frac{{{{(40)}^2}{{\sin }^2}\theta }}{{2 \times 9.8}}\) or \({\sin ^2}\theta = \frac{{25 \times 2 \times 9.8}}{{1600}}\) \( = 0.3063\) or \(\sin \theta = 0.5534\) \( = \sin 33.6^\circ \) or \(\theta = 33.6^\circ \) \(\therefore \) Horizontal range \((R) = \frac{{{u^2}\sin 2\theta }}{g}\) \( = \frac{{{{(40)}^2}\sin 2 \times 33.6^\circ }}{{9.8}}\) \( = \frac{{1600 \times \sin 67.2^\circ }}{{9.8}}\) \( = \frac{{1600 \times 0.9219}}{{9.8}} = 150.5\,m\)
NCERT Exemplar
PHXI04:MOTION IN A PLANE
362021
The range of the projectile projected at an angle of \(15^{\circ}\) with horizontal is \(50\;m\). If the projectile is projected with same velocity at an angle of \(45^{\circ}\) with horizontal, then its range will be
362017
Assertion : When \(\theta=45^{\circ}\) or \(135^{\circ}\), the value of \(R\) remains the same, only the sign changes. Reason : \(R = \frac{{{u^2}\sin 2\theta }}{g}\)
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.
Explanation:
Horizontal range, \(R=\dfrac{u^{2} \sin 2 \theta}{g}\) When \(\theta=45^{\circ}\), maximum horizontal range, \(R_{\max }=\dfrac{u^{2}}{g} \sin 90^{\circ}=\dfrac{u^{2}}{g}\) When \(\theta=135^{\circ}\), maximum horizontal range, \(R=\dfrac{u^{2}}{g} \sin 270^{\circ}=\dfrac{-u^{2}}{g}\) (Negative sign implies opposite direction)
AIIMS - 2017
PHXI04:MOTION IN A PLANE
362018
For an object thrown at \(45^\circ \) to horizontal, the maximum height (\(H\)) and horizontal range (\(R\)) are related as
362019
Statement A : When the range of projectile is maximum, the time of flight is the largest. Statement B : Range is maximum when angle of projection is \(45^\circ \)
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
\(T = \frac{{2u\sin \theta }}{g}\), it will maximum, when \(\theta = 0^\circ \). \({R_{\max }} = \frac{{{u^2}}}{g},{\rm{for}}\,\,\,\theta = 45^\circ \) So option (2) is correct.
PHXI04:MOTION IN A PLANE
362020
The ceiling of a long hall is 25\(m\) high. Then, the maximum horizontal distance that a ball thrown with a speed of 40 \(m\)/\(s\) can go without hitting the ceiling of the hall, is
1 \(95.5\,m\)
2 \(105.5\,m\)
3 \(100\,m\)
4 \(150.5\,\,m\)
Explanation:
Given, initial velocity \((u) = 40\,m/s\) Height of the hall \((H) = 25\,m\) Let the angle of projection of the ball be \(\theta \), when maximum height attained by it be 25\(m\). Maximum height attained by the ball \(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\) \(25 = \frac{{{{(40)}^2}{{\sin }^2}\theta }}{{2 \times 9.8}}\) or \({\sin ^2}\theta = \frac{{25 \times 2 \times 9.8}}{{1600}}\) \( = 0.3063\) or \(\sin \theta = 0.5534\) \( = \sin 33.6^\circ \) or \(\theta = 33.6^\circ \) \(\therefore \) Horizontal range \((R) = \frac{{{u^2}\sin 2\theta }}{g}\) \( = \frac{{{{(40)}^2}\sin 2 \times 33.6^\circ }}{{9.8}}\) \( = \frac{{1600 \times \sin 67.2^\circ }}{{9.8}}\) \( = \frac{{1600 \times 0.9219}}{{9.8}} = 150.5\,m\)
NCERT Exemplar
PHXI04:MOTION IN A PLANE
362021
The range of the projectile projected at an angle of \(15^{\circ}\) with horizontal is \(50\;m\). If the projectile is projected with same velocity at an angle of \(45^{\circ}\) with horizontal, then its range will be
362017
Assertion : When \(\theta=45^{\circ}\) or \(135^{\circ}\), the value of \(R\) remains the same, only the sign changes. Reason : \(R = \frac{{{u^2}\sin 2\theta }}{g}\)
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.
Explanation:
Horizontal range, \(R=\dfrac{u^{2} \sin 2 \theta}{g}\) When \(\theta=45^{\circ}\), maximum horizontal range, \(R_{\max }=\dfrac{u^{2}}{g} \sin 90^{\circ}=\dfrac{u^{2}}{g}\) When \(\theta=135^{\circ}\), maximum horizontal range, \(R=\dfrac{u^{2}}{g} \sin 270^{\circ}=\dfrac{-u^{2}}{g}\) (Negative sign implies opposite direction)
AIIMS - 2017
PHXI04:MOTION IN A PLANE
362018
For an object thrown at \(45^\circ \) to horizontal, the maximum height (\(H\)) and horizontal range (\(R\)) are related as
362019
Statement A : When the range of projectile is maximum, the time of flight is the largest. Statement B : Range is maximum when angle of projection is \(45^\circ \)
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
\(T = \frac{{2u\sin \theta }}{g}\), it will maximum, when \(\theta = 0^\circ \). \({R_{\max }} = \frac{{{u^2}}}{g},{\rm{for}}\,\,\,\theta = 45^\circ \) So option (2) is correct.
PHXI04:MOTION IN A PLANE
362020
The ceiling of a long hall is 25\(m\) high. Then, the maximum horizontal distance that a ball thrown with a speed of 40 \(m\)/\(s\) can go without hitting the ceiling of the hall, is
1 \(95.5\,m\)
2 \(105.5\,m\)
3 \(100\,m\)
4 \(150.5\,\,m\)
Explanation:
Given, initial velocity \((u) = 40\,m/s\) Height of the hall \((H) = 25\,m\) Let the angle of projection of the ball be \(\theta \), when maximum height attained by it be 25\(m\). Maximum height attained by the ball \(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\) \(25 = \frac{{{{(40)}^2}{{\sin }^2}\theta }}{{2 \times 9.8}}\) or \({\sin ^2}\theta = \frac{{25 \times 2 \times 9.8}}{{1600}}\) \( = 0.3063\) or \(\sin \theta = 0.5534\) \( = \sin 33.6^\circ \) or \(\theta = 33.6^\circ \) \(\therefore \) Horizontal range \((R) = \frac{{{u^2}\sin 2\theta }}{g}\) \( = \frac{{{{(40)}^2}\sin 2 \times 33.6^\circ }}{{9.8}}\) \( = \frac{{1600 \times \sin 67.2^\circ }}{{9.8}}\) \( = \frac{{1600 \times 0.9219}}{{9.8}} = 150.5\,m\)
NCERT Exemplar
PHXI04:MOTION IN A PLANE
362021
The range of the projectile projected at an angle of \(15^{\circ}\) with horizontal is \(50\;m\). If the projectile is projected with same velocity at an angle of \(45^{\circ}\) with horizontal, then its range will be