361978
Two bodies are thrown up at angles of \(45^{\circ}\) and \(60^{\circ}\) respectively, with the horizontal. If both bodies attain same vertical height, then ratio of velocities with which these are thrown is
1 \(\sqrt{2 / 3}\)
2 \(2 / \sqrt{3}\)
3 \(\sqrt{3 / 2}\)
4 \(\sqrt{3} / 2\)
Explanation:
According to the question, heights of both bodies are equal. \(\Rightarrow \dfrac{u_{1}^{2} \sin ^{2} \theta_{1}}{2 g}=\dfrac{u_{2}^{2} \sin ^{2} \theta_{2}}{2 g}\) or \(\dfrac{u_{1}}{u_{2}}=\dfrac{\sin \theta_{2}}{\sin \theta_{1}}=\dfrac{\sin 60^{\circ}}{\sin 45^{\circ}}=\dfrac{\sqrt{3} / 2}{1 / \sqrt{2}}=\sqrt{\dfrac{3}{2}}\)
PHXI04:MOTION IN A PLANE
361979
A ball is thrown from ground level so as to just clear a wall \(4\,m\) height at a distance of \(4\,m\) and falls at a distance of 14 \(m\) from the wall. The magnitude of velocity in \(m\)/\(s\) is.
1 \(13.6\,\)
2 \(7.8\)
3 \(18.5\)
4 \(12.5\)
Explanation:
We known the equation of projectile as \( \Rightarrow y = x\tan \theta \left[ {1 - \frac{x}{R}} \right]\) Where \(x\) and \(y\) can be any point on the projectile we can take \((x,y) = (4m,4m)\) \(4 = 4\tan \theta \left[ {1 - \frac{4}{{18}}} \right] \Rightarrow \tan \theta = \frac{9}{7}\) \(\sin \theta = 9/\sqrt {130} \) \(\cos \theta = 7/\sqrt {130} \) \(R = \frac{{{u^2}\sin 2\theta }}{g}\) \({u^2} = \frac{{gR}}{{2\sin \theta \cos \theta }} \Rightarrow u = 13.6\,m/s\)
PHXI04:MOTION IN A PLANE
361980
A stone is thrown at an angle \(\theta \) to the horizontal reaches a maximum height H. Then the time of flight of stone will be:
361981
A stone is thrown vertically at a speed of \(30\,m{s^{ - 1}}\) making an angle of \(45^\circ \) with the horizontal. What is the maximum height reached by the stone ? Take \(g = 10\,\,m{s^{ - 2}}\).
1 \(15\,m\)
2 \(30\,m\)
3 \(10\,m\)
4 \(22.5\,m\)
Explanation:
Maximum height reached by the stone is \(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\) Here, \(u = 30\,m{s^{ - 1}},\theta = 45^\circ ,g = 10\,m{s^{ - 1}}\) \(\therefore {H_{\max }} = \frac{{{{\left( {30} \right)}^2}{{\sin }^2}45}}{{2 \times 10}} = \frac{{30 \times 30 \times {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}{{2 \times 10}}\) \( = 22.5\,m\)
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PHXI04:MOTION IN A PLANE
361978
Two bodies are thrown up at angles of \(45^{\circ}\) and \(60^{\circ}\) respectively, with the horizontal. If both bodies attain same vertical height, then ratio of velocities with which these are thrown is
1 \(\sqrt{2 / 3}\)
2 \(2 / \sqrt{3}\)
3 \(\sqrt{3 / 2}\)
4 \(\sqrt{3} / 2\)
Explanation:
According to the question, heights of both bodies are equal. \(\Rightarrow \dfrac{u_{1}^{2} \sin ^{2} \theta_{1}}{2 g}=\dfrac{u_{2}^{2} \sin ^{2} \theta_{2}}{2 g}\) or \(\dfrac{u_{1}}{u_{2}}=\dfrac{\sin \theta_{2}}{\sin \theta_{1}}=\dfrac{\sin 60^{\circ}}{\sin 45^{\circ}}=\dfrac{\sqrt{3} / 2}{1 / \sqrt{2}}=\sqrt{\dfrac{3}{2}}\)
PHXI04:MOTION IN A PLANE
361979
A ball is thrown from ground level so as to just clear a wall \(4\,m\) height at a distance of \(4\,m\) and falls at a distance of 14 \(m\) from the wall. The magnitude of velocity in \(m\)/\(s\) is.
1 \(13.6\,\)
2 \(7.8\)
3 \(18.5\)
4 \(12.5\)
Explanation:
We known the equation of projectile as \( \Rightarrow y = x\tan \theta \left[ {1 - \frac{x}{R}} \right]\) Where \(x\) and \(y\) can be any point on the projectile we can take \((x,y) = (4m,4m)\) \(4 = 4\tan \theta \left[ {1 - \frac{4}{{18}}} \right] \Rightarrow \tan \theta = \frac{9}{7}\) \(\sin \theta = 9/\sqrt {130} \) \(\cos \theta = 7/\sqrt {130} \) \(R = \frac{{{u^2}\sin 2\theta }}{g}\) \({u^2} = \frac{{gR}}{{2\sin \theta \cos \theta }} \Rightarrow u = 13.6\,m/s\)
PHXI04:MOTION IN A PLANE
361980
A stone is thrown at an angle \(\theta \) to the horizontal reaches a maximum height H. Then the time of flight of stone will be:
361981
A stone is thrown vertically at a speed of \(30\,m{s^{ - 1}}\) making an angle of \(45^\circ \) with the horizontal. What is the maximum height reached by the stone ? Take \(g = 10\,\,m{s^{ - 2}}\).
1 \(15\,m\)
2 \(30\,m\)
3 \(10\,m\)
4 \(22.5\,m\)
Explanation:
Maximum height reached by the stone is \(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\) Here, \(u = 30\,m{s^{ - 1}},\theta = 45^\circ ,g = 10\,m{s^{ - 1}}\) \(\therefore {H_{\max }} = \frac{{{{\left( {30} \right)}^2}{{\sin }^2}45}}{{2 \times 10}} = \frac{{30 \times 30 \times {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}{{2 \times 10}}\) \( = 22.5\,m\)
361978
Two bodies are thrown up at angles of \(45^{\circ}\) and \(60^{\circ}\) respectively, with the horizontal. If both bodies attain same vertical height, then ratio of velocities with which these are thrown is
1 \(\sqrt{2 / 3}\)
2 \(2 / \sqrt{3}\)
3 \(\sqrt{3 / 2}\)
4 \(\sqrt{3} / 2\)
Explanation:
According to the question, heights of both bodies are equal. \(\Rightarrow \dfrac{u_{1}^{2} \sin ^{2} \theta_{1}}{2 g}=\dfrac{u_{2}^{2} \sin ^{2} \theta_{2}}{2 g}\) or \(\dfrac{u_{1}}{u_{2}}=\dfrac{\sin \theta_{2}}{\sin \theta_{1}}=\dfrac{\sin 60^{\circ}}{\sin 45^{\circ}}=\dfrac{\sqrt{3} / 2}{1 / \sqrt{2}}=\sqrt{\dfrac{3}{2}}\)
PHXI04:MOTION IN A PLANE
361979
A ball is thrown from ground level so as to just clear a wall \(4\,m\) height at a distance of \(4\,m\) and falls at a distance of 14 \(m\) from the wall. The magnitude of velocity in \(m\)/\(s\) is.
1 \(13.6\,\)
2 \(7.8\)
3 \(18.5\)
4 \(12.5\)
Explanation:
We known the equation of projectile as \( \Rightarrow y = x\tan \theta \left[ {1 - \frac{x}{R}} \right]\) Where \(x\) and \(y\) can be any point on the projectile we can take \((x,y) = (4m,4m)\) \(4 = 4\tan \theta \left[ {1 - \frac{4}{{18}}} \right] \Rightarrow \tan \theta = \frac{9}{7}\) \(\sin \theta = 9/\sqrt {130} \) \(\cos \theta = 7/\sqrt {130} \) \(R = \frac{{{u^2}\sin 2\theta }}{g}\) \({u^2} = \frac{{gR}}{{2\sin \theta \cos \theta }} \Rightarrow u = 13.6\,m/s\)
PHXI04:MOTION IN A PLANE
361980
A stone is thrown at an angle \(\theta \) to the horizontal reaches a maximum height H. Then the time of flight of stone will be:
361981
A stone is thrown vertically at a speed of \(30\,m{s^{ - 1}}\) making an angle of \(45^\circ \) with the horizontal. What is the maximum height reached by the stone ? Take \(g = 10\,\,m{s^{ - 2}}\).
1 \(15\,m\)
2 \(30\,m\)
3 \(10\,m\)
4 \(22.5\,m\)
Explanation:
Maximum height reached by the stone is \(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\) Here, \(u = 30\,m{s^{ - 1}},\theta = 45^\circ ,g = 10\,m{s^{ - 1}}\) \(\therefore {H_{\max }} = \frac{{{{\left( {30} \right)}^2}{{\sin }^2}45}}{{2 \times 10}} = \frac{{30 \times 30 \times {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}{{2 \times 10}}\) \( = 22.5\,m\)
361978
Two bodies are thrown up at angles of \(45^{\circ}\) and \(60^{\circ}\) respectively, with the horizontal. If both bodies attain same vertical height, then ratio of velocities with which these are thrown is
1 \(\sqrt{2 / 3}\)
2 \(2 / \sqrt{3}\)
3 \(\sqrt{3 / 2}\)
4 \(\sqrt{3} / 2\)
Explanation:
According to the question, heights of both bodies are equal. \(\Rightarrow \dfrac{u_{1}^{2} \sin ^{2} \theta_{1}}{2 g}=\dfrac{u_{2}^{2} \sin ^{2} \theta_{2}}{2 g}\) or \(\dfrac{u_{1}}{u_{2}}=\dfrac{\sin \theta_{2}}{\sin \theta_{1}}=\dfrac{\sin 60^{\circ}}{\sin 45^{\circ}}=\dfrac{\sqrt{3} / 2}{1 / \sqrt{2}}=\sqrt{\dfrac{3}{2}}\)
PHXI04:MOTION IN A PLANE
361979
A ball is thrown from ground level so as to just clear a wall \(4\,m\) height at a distance of \(4\,m\) and falls at a distance of 14 \(m\) from the wall. The magnitude of velocity in \(m\)/\(s\) is.
1 \(13.6\,\)
2 \(7.8\)
3 \(18.5\)
4 \(12.5\)
Explanation:
We known the equation of projectile as \( \Rightarrow y = x\tan \theta \left[ {1 - \frac{x}{R}} \right]\) Where \(x\) and \(y\) can be any point on the projectile we can take \((x,y) = (4m,4m)\) \(4 = 4\tan \theta \left[ {1 - \frac{4}{{18}}} \right] \Rightarrow \tan \theta = \frac{9}{7}\) \(\sin \theta = 9/\sqrt {130} \) \(\cos \theta = 7/\sqrt {130} \) \(R = \frac{{{u^2}\sin 2\theta }}{g}\) \({u^2} = \frac{{gR}}{{2\sin \theta \cos \theta }} \Rightarrow u = 13.6\,m/s\)
PHXI04:MOTION IN A PLANE
361980
A stone is thrown at an angle \(\theta \) to the horizontal reaches a maximum height H. Then the time of flight of stone will be:
361981
A stone is thrown vertically at a speed of \(30\,m{s^{ - 1}}\) making an angle of \(45^\circ \) with the horizontal. What is the maximum height reached by the stone ? Take \(g = 10\,\,m{s^{ - 2}}\).
1 \(15\,m\)
2 \(30\,m\)
3 \(10\,m\)
4 \(22.5\,m\)
Explanation:
Maximum height reached by the stone is \(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\) Here, \(u = 30\,m{s^{ - 1}},\theta = 45^\circ ,g = 10\,m{s^{ - 1}}\) \(\therefore {H_{\max }} = \frac{{{{\left( {30} \right)}^2}{{\sin }^2}45}}{{2 \times 10}} = \frac{{30 \times 30 \times {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}{{2 \times 10}}\) \( = 22.5\,m\)