NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in Plane
143784
The angle of projection of a projectile, for which the horizontal range and the maximum height are equal, is:
1 \(\tan ^{-1}(\sqrt{(3)})\)
2 \(\tan ^{-1}(4)\)
3 \(\tan ^{-1}(\sqrt{2})\)
4 \(\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
Explanation:
B We know that, Maximum height of projectile \((\mathrm{H})=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Range \((\mathrm{R})=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) According to question- \(\mathrm{H}=\mathrm{R}\) \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) \(\frac{\sin ^{2} \theta}{2}=2 \sin \theta \cos \theta\) \(\frac{\sin \theta}{2}=2 \cos \theta\) \(\tan \theta=4\) \(\theta=\tan ^{-1}(4)\)
APEAMCET(Medical)-1999
Motion in Plane
143785
The trajectory of a projectile projected from origin is given by the equation \(y=x-\frac{2 x^{2}}{5}\). The initial velocity of the projectile is:
1 \(\frac{2}{5} \mathrm{~ms}^{-1}\)
2 \(5 \mathrm{~ms}^{-1}\)
3 \(25 \mathrm{~ms}^{-1}\)
4 \(\frac{5}{2} \mathrm{~ms}^{-1}\)
Explanation:
B Given, \(y=x-\frac{2 x^{2}}{5}\) We know, Equation of projectile \((\mathrm{y})=\mathrm{x} \tan \theta-\frac{\mathrm{gx}^{2}}{2 \mathrm{u}^{2} \cos ^{2} \theta}\). Comparing the equation (i) and (ii), we get- \(\tan \theta=1\) \(\theta=45^{\circ}\) And, \(\frac{g}{2 u^{2} \cos ^{2} \theta}=\frac{2}{5}\) \(\frac{10}{2 u^{2} \cos ^{2} \theta}=\frac{2}{5}\) \(\frac{5}{u^{2} \cos ^{2} \theta}=\frac{2}{5}\) \(u^{2}=\frac{5 \times 5}{\left(\cos 45^{\circ}\right)^{2} \times 2}\) \(u^{2}=\frac{25}{\frac{1}{2} \times 2}\) \(u=5 \mathrm{~m} / \mathrm{s}\) \(\left[\therefore \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]\)
Karnataka CET-2019
Motion in Plane
143786
Three projectiles \(A, B\) and \(C\) are projected at an angle of \(30^{\circ}, 45^{\circ}, 60^{\circ}\) respectively. If \(R_{A}, R_{B}\) and \(R_{C}\) are ranges of \(A\), \(B\) and \(C\) respectively, then (velocity of projection is same for \(A, B\) and C) :
D Given, \(\theta_{\mathrm{A}}=30^{\circ}, \theta_{\mathrm{B}}=45^{\circ}, \theta_{\mathrm{C}}=60^{\circ}\) We know that, Range of projectile motion \((R)=\frac{u^{2} \sin 2 \theta}{g}\) \(\mathrm{R} \propto \sin 2 \theta \quad\{\mathrm{u} \text { is same for all }\}\) For, \(\quad \theta_{\mathrm{A}}=30^{\circ}\) \(\sin 2 \theta_{\mathrm{A}}=\sin 60^{\circ}=0.866\) For \(\quad \theta_{\mathrm{B}}=45^{\circ}\) \(\sin 2 \theta_{\mathrm{B}}=\sin 90^{\circ}=1\) For \(\quad \theta_{\mathrm{C}}=60^{\circ}\) \(\sin 2 \theta_{C}=\sin 120^{\circ}=0.866\) So, \(\quad \mathrm{R}_{\mathrm{A}}=\mathrm{R}_{\mathrm{C}} \lt \mathrm{R}_{\mathrm{B}}\)
Karnataka CET-2016
Motion in Plane
143787
A stone is thrown vertically at a speed of \(30 \mathrm{~ms}\) 1 taking an angle of \(45^{\circ}\) with the horizontal. What is the maximum height reached by the stone? Take \(g=10 \mathrm{~ms}^{-2}\).
1 \(30 \mathrm{~m}\)
2 \(22.5 \mathrm{~m}\)
3 \(15 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
B Given, \(\mathrm{u}=30 \mathrm{~ms}^{-1}, \theta=45^{0}, \mathrm{~g}=10 \mathrm{~ms}^{-2}\), maximum height \((\mathrm{H})=\) ? Maximum height of the projectile moving with velocity \(\mathrm{v}\) at an angle \(\theta\) is given by \(\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) \(\mathrm{H}=\frac{30^{2} \times \sin ^{2}\left(45^{\circ}\right)}{2 \times 10}\) \(\mathrm{H}=\frac{900}{20} \times\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{900 \times 1}{20 \times 2}\) \(\mathrm{H}=22.5 \mathrm{~m}\)
143784
The angle of projection of a projectile, for which the horizontal range and the maximum height are equal, is:
1 \(\tan ^{-1}(\sqrt{(3)})\)
2 \(\tan ^{-1}(4)\)
3 \(\tan ^{-1}(\sqrt{2})\)
4 \(\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
Explanation:
B We know that, Maximum height of projectile \((\mathrm{H})=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Range \((\mathrm{R})=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) According to question- \(\mathrm{H}=\mathrm{R}\) \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) \(\frac{\sin ^{2} \theta}{2}=2 \sin \theta \cos \theta\) \(\frac{\sin \theta}{2}=2 \cos \theta\) \(\tan \theta=4\) \(\theta=\tan ^{-1}(4)\)
APEAMCET(Medical)-1999
Motion in Plane
143785
The trajectory of a projectile projected from origin is given by the equation \(y=x-\frac{2 x^{2}}{5}\). The initial velocity of the projectile is:
1 \(\frac{2}{5} \mathrm{~ms}^{-1}\)
2 \(5 \mathrm{~ms}^{-1}\)
3 \(25 \mathrm{~ms}^{-1}\)
4 \(\frac{5}{2} \mathrm{~ms}^{-1}\)
Explanation:
B Given, \(y=x-\frac{2 x^{2}}{5}\) We know, Equation of projectile \((\mathrm{y})=\mathrm{x} \tan \theta-\frac{\mathrm{gx}^{2}}{2 \mathrm{u}^{2} \cos ^{2} \theta}\). Comparing the equation (i) and (ii), we get- \(\tan \theta=1\) \(\theta=45^{\circ}\) And, \(\frac{g}{2 u^{2} \cos ^{2} \theta}=\frac{2}{5}\) \(\frac{10}{2 u^{2} \cos ^{2} \theta}=\frac{2}{5}\) \(\frac{5}{u^{2} \cos ^{2} \theta}=\frac{2}{5}\) \(u^{2}=\frac{5 \times 5}{\left(\cos 45^{\circ}\right)^{2} \times 2}\) \(u^{2}=\frac{25}{\frac{1}{2} \times 2}\) \(u=5 \mathrm{~m} / \mathrm{s}\) \(\left[\therefore \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]\)
Karnataka CET-2019
Motion in Plane
143786
Three projectiles \(A, B\) and \(C\) are projected at an angle of \(30^{\circ}, 45^{\circ}, 60^{\circ}\) respectively. If \(R_{A}, R_{B}\) and \(R_{C}\) are ranges of \(A\), \(B\) and \(C\) respectively, then (velocity of projection is same for \(A, B\) and C) :
D Given, \(\theta_{\mathrm{A}}=30^{\circ}, \theta_{\mathrm{B}}=45^{\circ}, \theta_{\mathrm{C}}=60^{\circ}\) We know that, Range of projectile motion \((R)=\frac{u^{2} \sin 2 \theta}{g}\) \(\mathrm{R} \propto \sin 2 \theta \quad\{\mathrm{u} \text { is same for all }\}\) For, \(\quad \theta_{\mathrm{A}}=30^{\circ}\) \(\sin 2 \theta_{\mathrm{A}}=\sin 60^{\circ}=0.866\) For \(\quad \theta_{\mathrm{B}}=45^{\circ}\) \(\sin 2 \theta_{\mathrm{B}}=\sin 90^{\circ}=1\) For \(\quad \theta_{\mathrm{C}}=60^{\circ}\) \(\sin 2 \theta_{C}=\sin 120^{\circ}=0.866\) So, \(\quad \mathrm{R}_{\mathrm{A}}=\mathrm{R}_{\mathrm{C}} \lt \mathrm{R}_{\mathrm{B}}\)
Karnataka CET-2016
Motion in Plane
143787
A stone is thrown vertically at a speed of \(30 \mathrm{~ms}\) 1 taking an angle of \(45^{\circ}\) with the horizontal. What is the maximum height reached by the stone? Take \(g=10 \mathrm{~ms}^{-2}\).
1 \(30 \mathrm{~m}\)
2 \(22.5 \mathrm{~m}\)
3 \(15 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
B Given, \(\mathrm{u}=30 \mathrm{~ms}^{-1}, \theta=45^{0}, \mathrm{~g}=10 \mathrm{~ms}^{-2}\), maximum height \((\mathrm{H})=\) ? Maximum height of the projectile moving with velocity \(\mathrm{v}\) at an angle \(\theta\) is given by \(\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) \(\mathrm{H}=\frac{30^{2} \times \sin ^{2}\left(45^{\circ}\right)}{2 \times 10}\) \(\mathrm{H}=\frac{900}{20} \times\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{900 \times 1}{20 \times 2}\) \(\mathrm{H}=22.5 \mathrm{~m}\)
143784
The angle of projection of a projectile, for which the horizontal range and the maximum height are equal, is:
1 \(\tan ^{-1}(\sqrt{(3)})\)
2 \(\tan ^{-1}(4)\)
3 \(\tan ^{-1}(\sqrt{2})\)
4 \(\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
Explanation:
B We know that, Maximum height of projectile \((\mathrm{H})=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Range \((\mathrm{R})=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) According to question- \(\mathrm{H}=\mathrm{R}\) \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) \(\frac{\sin ^{2} \theta}{2}=2 \sin \theta \cos \theta\) \(\frac{\sin \theta}{2}=2 \cos \theta\) \(\tan \theta=4\) \(\theta=\tan ^{-1}(4)\)
APEAMCET(Medical)-1999
Motion in Plane
143785
The trajectory of a projectile projected from origin is given by the equation \(y=x-\frac{2 x^{2}}{5}\). The initial velocity of the projectile is:
1 \(\frac{2}{5} \mathrm{~ms}^{-1}\)
2 \(5 \mathrm{~ms}^{-1}\)
3 \(25 \mathrm{~ms}^{-1}\)
4 \(\frac{5}{2} \mathrm{~ms}^{-1}\)
Explanation:
B Given, \(y=x-\frac{2 x^{2}}{5}\) We know, Equation of projectile \((\mathrm{y})=\mathrm{x} \tan \theta-\frac{\mathrm{gx}^{2}}{2 \mathrm{u}^{2} \cos ^{2} \theta}\). Comparing the equation (i) and (ii), we get- \(\tan \theta=1\) \(\theta=45^{\circ}\) And, \(\frac{g}{2 u^{2} \cos ^{2} \theta}=\frac{2}{5}\) \(\frac{10}{2 u^{2} \cos ^{2} \theta}=\frac{2}{5}\) \(\frac{5}{u^{2} \cos ^{2} \theta}=\frac{2}{5}\) \(u^{2}=\frac{5 \times 5}{\left(\cos 45^{\circ}\right)^{2} \times 2}\) \(u^{2}=\frac{25}{\frac{1}{2} \times 2}\) \(u=5 \mathrm{~m} / \mathrm{s}\) \(\left[\therefore \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]\)
Karnataka CET-2019
Motion in Plane
143786
Three projectiles \(A, B\) and \(C\) are projected at an angle of \(30^{\circ}, 45^{\circ}, 60^{\circ}\) respectively. If \(R_{A}, R_{B}\) and \(R_{C}\) are ranges of \(A\), \(B\) and \(C\) respectively, then (velocity of projection is same for \(A, B\) and C) :
D Given, \(\theta_{\mathrm{A}}=30^{\circ}, \theta_{\mathrm{B}}=45^{\circ}, \theta_{\mathrm{C}}=60^{\circ}\) We know that, Range of projectile motion \((R)=\frac{u^{2} \sin 2 \theta}{g}\) \(\mathrm{R} \propto \sin 2 \theta \quad\{\mathrm{u} \text { is same for all }\}\) For, \(\quad \theta_{\mathrm{A}}=30^{\circ}\) \(\sin 2 \theta_{\mathrm{A}}=\sin 60^{\circ}=0.866\) For \(\quad \theta_{\mathrm{B}}=45^{\circ}\) \(\sin 2 \theta_{\mathrm{B}}=\sin 90^{\circ}=1\) For \(\quad \theta_{\mathrm{C}}=60^{\circ}\) \(\sin 2 \theta_{C}=\sin 120^{\circ}=0.866\) So, \(\quad \mathrm{R}_{\mathrm{A}}=\mathrm{R}_{\mathrm{C}} \lt \mathrm{R}_{\mathrm{B}}\)
Karnataka CET-2016
Motion in Plane
143787
A stone is thrown vertically at a speed of \(30 \mathrm{~ms}\) 1 taking an angle of \(45^{\circ}\) with the horizontal. What is the maximum height reached by the stone? Take \(g=10 \mathrm{~ms}^{-2}\).
1 \(30 \mathrm{~m}\)
2 \(22.5 \mathrm{~m}\)
3 \(15 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
B Given, \(\mathrm{u}=30 \mathrm{~ms}^{-1}, \theta=45^{0}, \mathrm{~g}=10 \mathrm{~ms}^{-2}\), maximum height \((\mathrm{H})=\) ? Maximum height of the projectile moving with velocity \(\mathrm{v}\) at an angle \(\theta\) is given by \(\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) \(\mathrm{H}=\frac{30^{2} \times \sin ^{2}\left(45^{\circ}\right)}{2 \times 10}\) \(\mathrm{H}=\frac{900}{20} \times\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{900 \times 1}{20 \times 2}\) \(\mathrm{H}=22.5 \mathrm{~m}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Motion in Plane
143784
The angle of projection of a projectile, for which the horizontal range and the maximum height are equal, is:
1 \(\tan ^{-1}(\sqrt{(3)})\)
2 \(\tan ^{-1}(4)\)
3 \(\tan ^{-1}(\sqrt{2})\)
4 \(\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
Explanation:
B We know that, Maximum height of projectile \((\mathrm{H})=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) Range \((\mathrm{R})=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) According to question- \(\mathrm{H}=\mathrm{R}\) \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) \(\frac{\sin ^{2} \theta}{2}=2 \sin \theta \cos \theta\) \(\frac{\sin \theta}{2}=2 \cos \theta\) \(\tan \theta=4\) \(\theta=\tan ^{-1}(4)\)
APEAMCET(Medical)-1999
Motion in Plane
143785
The trajectory of a projectile projected from origin is given by the equation \(y=x-\frac{2 x^{2}}{5}\). The initial velocity of the projectile is:
1 \(\frac{2}{5} \mathrm{~ms}^{-1}\)
2 \(5 \mathrm{~ms}^{-1}\)
3 \(25 \mathrm{~ms}^{-1}\)
4 \(\frac{5}{2} \mathrm{~ms}^{-1}\)
Explanation:
B Given, \(y=x-\frac{2 x^{2}}{5}\) We know, Equation of projectile \((\mathrm{y})=\mathrm{x} \tan \theta-\frac{\mathrm{gx}^{2}}{2 \mathrm{u}^{2} \cos ^{2} \theta}\). Comparing the equation (i) and (ii), we get- \(\tan \theta=1\) \(\theta=45^{\circ}\) And, \(\frac{g}{2 u^{2} \cos ^{2} \theta}=\frac{2}{5}\) \(\frac{10}{2 u^{2} \cos ^{2} \theta}=\frac{2}{5}\) \(\frac{5}{u^{2} \cos ^{2} \theta}=\frac{2}{5}\) \(u^{2}=\frac{5 \times 5}{\left(\cos 45^{\circ}\right)^{2} \times 2}\) \(u^{2}=\frac{25}{\frac{1}{2} \times 2}\) \(u=5 \mathrm{~m} / \mathrm{s}\) \(\left[\therefore \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]\)
Karnataka CET-2019
Motion in Plane
143786
Three projectiles \(A, B\) and \(C\) are projected at an angle of \(30^{\circ}, 45^{\circ}, 60^{\circ}\) respectively. If \(R_{A}, R_{B}\) and \(R_{C}\) are ranges of \(A\), \(B\) and \(C\) respectively, then (velocity of projection is same for \(A, B\) and C) :
D Given, \(\theta_{\mathrm{A}}=30^{\circ}, \theta_{\mathrm{B}}=45^{\circ}, \theta_{\mathrm{C}}=60^{\circ}\) We know that, Range of projectile motion \((R)=\frac{u^{2} \sin 2 \theta}{g}\) \(\mathrm{R} \propto \sin 2 \theta \quad\{\mathrm{u} \text { is same for all }\}\) For, \(\quad \theta_{\mathrm{A}}=30^{\circ}\) \(\sin 2 \theta_{\mathrm{A}}=\sin 60^{\circ}=0.866\) For \(\quad \theta_{\mathrm{B}}=45^{\circ}\) \(\sin 2 \theta_{\mathrm{B}}=\sin 90^{\circ}=1\) For \(\quad \theta_{\mathrm{C}}=60^{\circ}\) \(\sin 2 \theta_{C}=\sin 120^{\circ}=0.866\) So, \(\quad \mathrm{R}_{\mathrm{A}}=\mathrm{R}_{\mathrm{C}} \lt \mathrm{R}_{\mathrm{B}}\)
Karnataka CET-2016
Motion in Plane
143787
A stone is thrown vertically at a speed of \(30 \mathrm{~ms}\) 1 taking an angle of \(45^{\circ}\) with the horizontal. What is the maximum height reached by the stone? Take \(g=10 \mathrm{~ms}^{-2}\).
1 \(30 \mathrm{~m}\)
2 \(22.5 \mathrm{~m}\)
3 \(15 \mathrm{~m}\)
4 \(10 \mathrm{~m}\)
Explanation:
B Given, \(\mathrm{u}=30 \mathrm{~ms}^{-1}, \theta=45^{0}, \mathrm{~g}=10 \mathrm{~ms}^{-2}\), maximum height \((\mathrm{H})=\) ? Maximum height of the projectile moving with velocity \(\mathrm{v}\) at an angle \(\theta\) is given by \(\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\) \(\mathrm{H}=\frac{30^{2} \times \sin ^{2}\left(45^{\circ}\right)}{2 \times 10}\) \(\mathrm{H}=\frac{900}{20} \times\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{900 \times 1}{20 \times 2}\) \(\mathrm{H}=22.5 \mathrm{~m}\)