269922
A cricket ball is hit for a six leaving the bat at an angle of\(60^{\circ}\) to the horizontal with kinetic energy ' \(k\) '. At the top, K.E. of the ball is
1 Zero
2 \(\mathrm{k}\)
3 \(\frac{\mathrm{k}}{4}\)
4 \(\frac{\mathrm{k}}{\sqrt{2}}\)
Explanation:
\(H_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}\)
JEE MAIN-2007
Motion in Plane
269876
A particle is projected in \(x y\) plane with\(y\)-axis along vertical, the point of projection is origin. The equation of the path is \(y=\sqrt{3} x-\frac{g}{2} x^{2}\). where \(y\) and \(x\) are in \(m\). Then the speed of projection in \(\mathrm{ms}^{-1}\) is
1 2
2 \(\sqrt{3}\)
3 4
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
Compare the equation with
\(y=x \tan \theta-\frac{g}{2 u^{2} \cos ^{2} \theta} x^{2}\)
Motion in Plane
269877
If a body is thrown with a speed of\(19.6 \mathrm{~m} / \mathrm{s}\) making an angle of \(30^{\circ}\) with the horizontal, then the time of flight is
1 \(1 \mathrm{~s}\)
2 \(2 \mathrm{~s}\)
3 \(2 \sqrt{3} \mathrm{~s}\)
4 \(5 \mathrm{~s}\)
Explanation:
\(\mathrm{T}=\frac{2 u \sin \theta}{g}\)
Motion in Plane
269878
A particle is projected with an initial velocity of\(200 \mathrm{~m} / \mathrm{s}\) in a direction making an angle of \(30^{\circ}\) with the vertical. The horizontal distance covered by the particle in \(3 s\) is
1 \(300 \mathrm{~m}\)
2 \(150 \mathrm{~m}\)
3 \(175 \mathrm{~m}\)
4 \(125 \mathrm{~m}\)
Explanation:
x \(=(u \cos \theta) t\)
Motion in Plane
269879
A body is projected with an initial velocity\(20 \mathrm{~m} / \mathrm{s}\) at \(60^{\circ}\) to the horizontal. Its initial velocity vector is __ \(\left(\mathbf{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
269922
A cricket ball is hit for a six leaving the bat at an angle of\(60^{\circ}\) to the horizontal with kinetic energy ' \(k\) '. At the top, K.E. of the ball is
1 Zero
2 \(\mathrm{k}\)
3 \(\frac{\mathrm{k}}{4}\)
4 \(\frac{\mathrm{k}}{\sqrt{2}}\)
Explanation:
\(H_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}\)
JEE MAIN-2007
Motion in Plane
269876
A particle is projected in \(x y\) plane with\(y\)-axis along vertical, the point of projection is origin. The equation of the path is \(y=\sqrt{3} x-\frac{g}{2} x^{2}\). where \(y\) and \(x\) are in \(m\). Then the speed of projection in \(\mathrm{ms}^{-1}\) is
1 2
2 \(\sqrt{3}\)
3 4
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
Compare the equation with
\(y=x \tan \theta-\frac{g}{2 u^{2} \cos ^{2} \theta} x^{2}\)
Motion in Plane
269877
If a body is thrown with a speed of\(19.6 \mathrm{~m} / \mathrm{s}\) making an angle of \(30^{\circ}\) with the horizontal, then the time of flight is
1 \(1 \mathrm{~s}\)
2 \(2 \mathrm{~s}\)
3 \(2 \sqrt{3} \mathrm{~s}\)
4 \(5 \mathrm{~s}\)
Explanation:
\(\mathrm{T}=\frac{2 u \sin \theta}{g}\)
Motion in Plane
269878
A particle is projected with an initial velocity of\(200 \mathrm{~m} / \mathrm{s}\) in a direction making an angle of \(30^{\circ}\) with the vertical. The horizontal distance covered by the particle in \(3 s\) is
1 \(300 \mathrm{~m}\)
2 \(150 \mathrm{~m}\)
3 \(175 \mathrm{~m}\)
4 \(125 \mathrm{~m}\)
Explanation:
x \(=(u \cos \theta) t\)
Motion in Plane
269879
A body is projected with an initial velocity\(20 \mathrm{~m} / \mathrm{s}\) at \(60^{\circ}\) to the horizontal. Its initial velocity vector is __ \(\left(\mathbf{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
269922
A cricket ball is hit for a six leaving the bat at an angle of\(60^{\circ}\) to the horizontal with kinetic energy ' \(k\) '. At the top, K.E. of the ball is
1 Zero
2 \(\mathrm{k}\)
3 \(\frac{\mathrm{k}}{4}\)
4 \(\frac{\mathrm{k}}{\sqrt{2}}\)
Explanation:
\(H_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}\)
JEE MAIN-2007
Motion in Plane
269876
A particle is projected in \(x y\) plane with\(y\)-axis along vertical, the point of projection is origin. The equation of the path is \(y=\sqrt{3} x-\frac{g}{2} x^{2}\). where \(y\) and \(x\) are in \(m\). Then the speed of projection in \(\mathrm{ms}^{-1}\) is
1 2
2 \(\sqrt{3}\)
3 4
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
Compare the equation with
\(y=x \tan \theta-\frac{g}{2 u^{2} \cos ^{2} \theta} x^{2}\)
Motion in Plane
269877
If a body is thrown with a speed of\(19.6 \mathrm{~m} / \mathrm{s}\) making an angle of \(30^{\circ}\) with the horizontal, then the time of flight is
1 \(1 \mathrm{~s}\)
2 \(2 \mathrm{~s}\)
3 \(2 \sqrt{3} \mathrm{~s}\)
4 \(5 \mathrm{~s}\)
Explanation:
\(\mathrm{T}=\frac{2 u \sin \theta}{g}\)
Motion in Plane
269878
A particle is projected with an initial velocity of\(200 \mathrm{~m} / \mathrm{s}\) in a direction making an angle of \(30^{\circ}\) with the vertical. The horizontal distance covered by the particle in \(3 s\) is
1 \(300 \mathrm{~m}\)
2 \(150 \mathrm{~m}\)
3 \(175 \mathrm{~m}\)
4 \(125 \mathrm{~m}\)
Explanation:
x \(=(u \cos \theta) t\)
Motion in Plane
269879
A body is projected with an initial velocity\(20 \mathrm{~m} / \mathrm{s}\) at \(60^{\circ}\) to the horizontal. Its initial velocity vector is __ \(\left(\mathbf{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
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Motion in Plane
269922
A cricket ball is hit for a six leaving the bat at an angle of\(60^{\circ}\) to the horizontal with kinetic energy ' \(k\) '. At the top, K.E. of the ball is
1 Zero
2 \(\mathrm{k}\)
3 \(\frac{\mathrm{k}}{4}\)
4 \(\frac{\mathrm{k}}{\sqrt{2}}\)
Explanation:
\(H_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}\)
JEE MAIN-2007
Motion in Plane
269876
A particle is projected in \(x y\) plane with\(y\)-axis along vertical, the point of projection is origin. The equation of the path is \(y=\sqrt{3} x-\frac{g}{2} x^{2}\). where \(y\) and \(x\) are in \(m\). Then the speed of projection in \(\mathrm{ms}^{-1}\) is
1 2
2 \(\sqrt{3}\)
3 4
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
Compare the equation with
\(y=x \tan \theta-\frac{g}{2 u^{2} \cos ^{2} \theta} x^{2}\)
Motion in Plane
269877
If a body is thrown with a speed of\(19.6 \mathrm{~m} / \mathrm{s}\) making an angle of \(30^{\circ}\) with the horizontal, then the time of flight is
1 \(1 \mathrm{~s}\)
2 \(2 \mathrm{~s}\)
3 \(2 \sqrt{3} \mathrm{~s}\)
4 \(5 \mathrm{~s}\)
Explanation:
\(\mathrm{T}=\frac{2 u \sin \theta}{g}\)
Motion in Plane
269878
A particle is projected with an initial velocity of\(200 \mathrm{~m} / \mathrm{s}\) in a direction making an angle of \(30^{\circ}\) with the vertical. The horizontal distance covered by the particle in \(3 s\) is
1 \(300 \mathrm{~m}\)
2 \(150 \mathrm{~m}\)
3 \(175 \mathrm{~m}\)
4 \(125 \mathrm{~m}\)
Explanation:
x \(=(u \cos \theta) t\)
Motion in Plane
269879
A body is projected with an initial velocity\(20 \mathrm{~m} / \mathrm{s}\) at \(60^{\circ}\) to the horizontal. Its initial velocity vector is __ \(\left(\mathbf{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
269922
A cricket ball is hit for a six leaving the bat at an angle of\(60^{\circ}\) to the horizontal with kinetic energy ' \(k\) '. At the top, K.E. of the ball is
1 Zero
2 \(\mathrm{k}\)
3 \(\frac{\mathrm{k}}{4}\)
4 \(\frac{\mathrm{k}}{\sqrt{2}}\)
Explanation:
\(H_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}\)
JEE MAIN-2007
Motion in Plane
269876
A particle is projected in \(x y\) plane with\(y\)-axis along vertical, the point of projection is origin. The equation of the path is \(y=\sqrt{3} x-\frac{g}{2} x^{2}\). where \(y\) and \(x\) are in \(m\). Then the speed of projection in \(\mathrm{ms}^{-1}\) is
1 2
2 \(\sqrt{3}\)
3 4
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
Compare the equation with
\(y=x \tan \theta-\frac{g}{2 u^{2} \cos ^{2} \theta} x^{2}\)
Motion in Plane
269877
If a body is thrown with a speed of\(19.6 \mathrm{~m} / \mathrm{s}\) making an angle of \(30^{\circ}\) with the horizontal, then the time of flight is
1 \(1 \mathrm{~s}\)
2 \(2 \mathrm{~s}\)
3 \(2 \sqrt{3} \mathrm{~s}\)
4 \(5 \mathrm{~s}\)
Explanation:
\(\mathrm{T}=\frac{2 u \sin \theta}{g}\)
Motion in Plane
269878
A particle is projected with an initial velocity of\(200 \mathrm{~m} / \mathrm{s}\) in a direction making an angle of \(30^{\circ}\) with the vertical. The horizontal distance covered by the particle in \(3 s\) is
1 \(300 \mathrm{~m}\)
2 \(150 \mathrm{~m}\)
3 \(175 \mathrm{~m}\)
4 \(125 \mathrm{~m}\)
Explanation:
x \(=(u \cos \theta) t\)
Motion in Plane
269879
A body is projected with an initial velocity\(20 \mathrm{~m} / \mathrm{s}\) at \(60^{\circ}\) to the horizontal. Its initial velocity vector is __ \(\left(\mathbf{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)