269880
A body is projected at an angle of\(30^{\circ}\) with the horizontal with momentum P. At its highest point the magnitude of the momentum is:
1 \(\frac{\sqrt{3}}{2} P\)
2 \(\frac{2}{\sqrt{3}} P\)
3 \(P\)
4 \(\frac{P}{2}\)
Explanation:
\(P_{x}=P \cos \theta, P_{y}=0\)
Motion in Plane
269881
The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is\(20 \mathrm{~ms}^{-1}\), its time of flight is \(\left(\mathrm{g}=\mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(2 \mathrm{~s}\)
2 \(2 \sqrt{2} s\)
3 \(\frac{1}{2} s\)
4 \(\frac{1}{\sqrt{2}} s\)
Explanation:
\(\quad P . E=K . E \Rightarrow \frac{1}{2} m u^{2} \sin ^{2} \theta=\frac{1}{2} m u^{2} \cos ^{2} \theta\) use \(T=\frac{2 u \sin \theta}{g}\)
Motion in Plane
269882
From a point on the ground a particle is projected with initial velocity\(u\), such that its horizontal range is maximum. The magnitude of average velocity during its ascent.
269883
The horizontal and vertical displacements of a projectile are given as\(\mathrm{x}=\mathrm{at}\) \& \(\mathrm{y}=b t-c t^{2}\). Then velocity of projection is
1 \(\sqrt{a^{2}+b^{2}}\)
2 \(\sqrt{b^{2}+c^{2}}\)
3 \(\sqrt{a^{2}+c^{2}}\)
4 \(\sqrt{\mathrm{b}^{2}-\mathrm{c}^{2}}\)
Explanation:
\(u_{x}=\frac{d x}{d t}, u_{y}=\frac{d y}{d t}\) and \(u=\sqrt{u_{x}^{2}+u_{y}^{2}}\)
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Motion in Plane
269880
A body is projected at an angle of\(30^{\circ}\) with the horizontal with momentum P. At its highest point the magnitude of the momentum is:
1 \(\frac{\sqrt{3}}{2} P\)
2 \(\frac{2}{\sqrt{3}} P\)
3 \(P\)
4 \(\frac{P}{2}\)
Explanation:
\(P_{x}=P \cos \theta, P_{y}=0\)
Motion in Plane
269881
The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is\(20 \mathrm{~ms}^{-1}\), its time of flight is \(\left(\mathrm{g}=\mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(2 \mathrm{~s}\)
2 \(2 \sqrt{2} s\)
3 \(\frac{1}{2} s\)
4 \(\frac{1}{\sqrt{2}} s\)
Explanation:
\(\quad P . E=K . E \Rightarrow \frac{1}{2} m u^{2} \sin ^{2} \theta=\frac{1}{2} m u^{2} \cos ^{2} \theta\) use \(T=\frac{2 u \sin \theta}{g}\)
Motion in Plane
269882
From a point on the ground a particle is projected with initial velocity\(u\), such that its horizontal range is maximum. The magnitude of average velocity during its ascent.
269883
The horizontal and vertical displacements of a projectile are given as\(\mathrm{x}=\mathrm{at}\) \& \(\mathrm{y}=b t-c t^{2}\). Then velocity of projection is
1 \(\sqrt{a^{2}+b^{2}}\)
2 \(\sqrt{b^{2}+c^{2}}\)
3 \(\sqrt{a^{2}+c^{2}}\)
4 \(\sqrt{\mathrm{b}^{2}-\mathrm{c}^{2}}\)
Explanation:
\(u_{x}=\frac{d x}{d t}, u_{y}=\frac{d y}{d t}\) and \(u=\sqrt{u_{x}^{2}+u_{y}^{2}}\)
269880
A body is projected at an angle of\(30^{\circ}\) with the horizontal with momentum P. At its highest point the magnitude of the momentum is:
1 \(\frac{\sqrt{3}}{2} P\)
2 \(\frac{2}{\sqrt{3}} P\)
3 \(P\)
4 \(\frac{P}{2}\)
Explanation:
\(P_{x}=P \cos \theta, P_{y}=0\)
Motion in Plane
269881
The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is\(20 \mathrm{~ms}^{-1}\), its time of flight is \(\left(\mathrm{g}=\mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(2 \mathrm{~s}\)
2 \(2 \sqrt{2} s\)
3 \(\frac{1}{2} s\)
4 \(\frac{1}{\sqrt{2}} s\)
Explanation:
\(\quad P . E=K . E \Rightarrow \frac{1}{2} m u^{2} \sin ^{2} \theta=\frac{1}{2} m u^{2} \cos ^{2} \theta\) use \(T=\frac{2 u \sin \theta}{g}\)
Motion in Plane
269882
From a point on the ground a particle is projected with initial velocity\(u\), such that its horizontal range is maximum. The magnitude of average velocity during its ascent.
269883
The horizontal and vertical displacements of a projectile are given as\(\mathrm{x}=\mathrm{at}\) \& \(\mathrm{y}=b t-c t^{2}\). Then velocity of projection is
1 \(\sqrt{a^{2}+b^{2}}\)
2 \(\sqrt{b^{2}+c^{2}}\)
3 \(\sqrt{a^{2}+c^{2}}\)
4 \(\sqrt{\mathrm{b}^{2}-\mathrm{c}^{2}}\)
Explanation:
\(u_{x}=\frac{d x}{d t}, u_{y}=\frac{d y}{d t}\) and \(u=\sqrt{u_{x}^{2}+u_{y}^{2}}\)
269880
A body is projected at an angle of\(30^{\circ}\) with the horizontal with momentum P. At its highest point the magnitude of the momentum is:
1 \(\frac{\sqrt{3}}{2} P\)
2 \(\frac{2}{\sqrt{3}} P\)
3 \(P\)
4 \(\frac{P}{2}\)
Explanation:
\(P_{x}=P \cos \theta, P_{y}=0\)
Motion in Plane
269881
The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is\(20 \mathrm{~ms}^{-1}\), its time of flight is \(\left(\mathrm{g}=\mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(2 \mathrm{~s}\)
2 \(2 \sqrt{2} s\)
3 \(\frac{1}{2} s\)
4 \(\frac{1}{\sqrt{2}} s\)
Explanation:
\(\quad P . E=K . E \Rightarrow \frac{1}{2} m u^{2} \sin ^{2} \theta=\frac{1}{2} m u^{2} \cos ^{2} \theta\) use \(T=\frac{2 u \sin \theta}{g}\)
Motion in Plane
269882
From a point on the ground a particle is projected with initial velocity\(u\), such that its horizontal range is maximum. The magnitude of average velocity during its ascent.
269883
The horizontal and vertical displacements of a projectile are given as\(\mathrm{x}=\mathrm{at}\) \& \(\mathrm{y}=b t-c t^{2}\). Then velocity of projection is
1 \(\sqrt{a^{2}+b^{2}}\)
2 \(\sqrt{b^{2}+c^{2}}\)
3 \(\sqrt{a^{2}+c^{2}}\)
4 \(\sqrt{\mathrm{b}^{2}-\mathrm{c}^{2}}\)
Explanation:
\(u_{x}=\frac{d x}{d t}, u_{y}=\frac{d y}{d t}\) and \(u=\sqrt{u_{x}^{2}+u_{y}^{2}}\)