143799
Two stones are projected with the same magnitude of velocity, but making different angles with horizontal, the angle of projection of one is \(\pi / 3\) and its maximum height is \(Y\), the maximum height attained by the other stone with \(\pi / 6\) angle of projection is
1 \(\mathrm{Y}\)
2 \(2 \mathrm{Y}\)
3 \(3 \mathrm{Y}\)
4 \(\frac{Y}{3}\)
Explanation:
D Given, \(\theta_{1}=\pi / 3, \mathrm{~h}_{1}=\mathrm{Y}, \theta_{2}=\pi / 6\) Let \(h_{2}\) be the maximum height of other stone. We know, \(h=\frac{u^{2} \sin ^{2} \theta}{2 g}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\sin ^{2} \theta_{1}}{\sin ^{2} \theta_{2}}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\sin ^{2}(\pi / 3)}{\sin ^{2}(\pi / 6)}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{(\sqrt{3} / 2)^{2}}{(1 / 2)^{2}}=\frac{3 / 4}{1 / 4}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{3}{1}\) \(\mathrm{h}_{2}=\frac{\mathrm{h}_{1}}{3} \quad\left[\mathrm{~h}_{1}=\mathrm{Y}\right]\) \(\therefore \quad \mathrm{h}_{2}=\frac{\mathrm{Y}}{3}\)
J and K-CET-2000
Motion in Plane
143802
A projectile is thrown with initial velocity \(v_{0}\) and angle \(30^{\circ}\) with the horizontal. If it remains in the air for \(1 \mathrm{sec}\), what was its initial velocity?
1 \(19.6 \mathrm{~m} / \mathrm{s}\)
2 \(9.8 \mathrm{~m} / \mathrm{s}\)
3 \(4.9 \mathrm{~m} / \mathrm{s}\)
4 \(1 \mathrm{~m} / \mathrm{s}\)
Explanation:
B As we know that time of flight in projectile motion is given by \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\) It is given that, \(\mathrm{T}=1 \mathrm{sec}, \quad \theta=30^{\circ}\) \(\mathrm{u}=\mathrm{u}_{0}(\text { initial velocity })\) \(\text { Take } \mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) Time of flight in projectile motion \(1=\frac{2 \mathrm{u}_{0} \sin 30^{\circ}}{9.8} \Rightarrow \mathrm{v}_{0}=\frac{9.8}{2 \times \sin 30^{\circ}}\) \(\mathrm{v}_{0}=9.8 \mathrm{~m} / \mathrm{s}\)
J and K-CET-2013
Motion in Plane
143803
A bird flies at an angle of \(60^{\circ}\) to the horizontal Its horizontal component of velocity is \(10 \mathrm{~m} \mathrm{~s}^{-1}\) . Find the vertical component of velocity in \(\mathrm{m}\) \(\mathbf{s}^{-1}\)
143804
The angle which the velocity vector of a projectile thrown with a velocity \(v\) at an angle \(\theta\) to the horizontal will make with the horizontal after time \(t\) of its being thrown up is:
D As in given question the graph of projectile will be given as Vertical velocity of projectile after time \(\mathrm{t}\) is \(v \sin \theta-\mathrm{gt} \quad\) (from equation of motion) Angle at any time \(t\), \(\tan \phi=\frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}}=\frac{\mathrm{v} \sin \theta-\mathrm{gt}}{\mathrm{v} \cos \theta}\) \(\phi=\tan ^{-1}\left(\frac{\mathrm{v} \sin \theta-\mathrm{gt}}{\mathrm{v} \cos \theta}\right)\)
143799
Two stones are projected with the same magnitude of velocity, but making different angles with horizontal, the angle of projection of one is \(\pi / 3\) and its maximum height is \(Y\), the maximum height attained by the other stone with \(\pi / 6\) angle of projection is
1 \(\mathrm{Y}\)
2 \(2 \mathrm{Y}\)
3 \(3 \mathrm{Y}\)
4 \(\frac{Y}{3}\)
Explanation:
D Given, \(\theta_{1}=\pi / 3, \mathrm{~h}_{1}=\mathrm{Y}, \theta_{2}=\pi / 6\) Let \(h_{2}\) be the maximum height of other stone. We know, \(h=\frac{u^{2} \sin ^{2} \theta}{2 g}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\sin ^{2} \theta_{1}}{\sin ^{2} \theta_{2}}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\sin ^{2}(\pi / 3)}{\sin ^{2}(\pi / 6)}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{(\sqrt{3} / 2)^{2}}{(1 / 2)^{2}}=\frac{3 / 4}{1 / 4}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{3}{1}\) \(\mathrm{h}_{2}=\frac{\mathrm{h}_{1}}{3} \quad\left[\mathrm{~h}_{1}=\mathrm{Y}\right]\) \(\therefore \quad \mathrm{h}_{2}=\frac{\mathrm{Y}}{3}\)
J and K-CET-2000
Motion in Plane
143802
A projectile is thrown with initial velocity \(v_{0}\) and angle \(30^{\circ}\) with the horizontal. If it remains in the air for \(1 \mathrm{sec}\), what was its initial velocity?
1 \(19.6 \mathrm{~m} / \mathrm{s}\)
2 \(9.8 \mathrm{~m} / \mathrm{s}\)
3 \(4.9 \mathrm{~m} / \mathrm{s}\)
4 \(1 \mathrm{~m} / \mathrm{s}\)
Explanation:
B As we know that time of flight in projectile motion is given by \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\) It is given that, \(\mathrm{T}=1 \mathrm{sec}, \quad \theta=30^{\circ}\) \(\mathrm{u}=\mathrm{u}_{0}(\text { initial velocity })\) \(\text { Take } \mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) Time of flight in projectile motion \(1=\frac{2 \mathrm{u}_{0} \sin 30^{\circ}}{9.8} \Rightarrow \mathrm{v}_{0}=\frac{9.8}{2 \times \sin 30^{\circ}}\) \(\mathrm{v}_{0}=9.8 \mathrm{~m} / \mathrm{s}\)
J and K-CET-2013
Motion in Plane
143803
A bird flies at an angle of \(60^{\circ}\) to the horizontal Its horizontal component of velocity is \(10 \mathrm{~m} \mathrm{~s}^{-1}\) . Find the vertical component of velocity in \(\mathrm{m}\) \(\mathbf{s}^{-1}\)
143804
The angle which the velocity vector of a projectile thrown with a velocity \(v\) at an angle \(\theta\) to the horizontal will make with the horizontal after time \(t\) of its being thrown up is:
D As in given question the graph of projectile will be given as Vertical velocity of projectile after time \(\mathrm{t}\) is \(v \sin \theta-\mathrm{gt} \quad\) (from equation of motion) Angle at any time \(t\), \(\tan \phi=\frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}}=\frac{\mathrm{v} \sin \theta-\mathrm{gt}}{\mathrm{v} \cos \theta}\) \(\phi=\tan ^{-1}\left(\frac{\mathrm{v} \sin \theta-\mathrm{gt}}{\mathrm{v} \cos \theta}\right)\)
143799
Two stones are projected with the same magnitude of velocity, but making different angles with horizontal, the angle of projection of one is \(\pi / 3\) and its maximum height is \(Y\), the maximum height attained by the other stone with \(\pi / 6\) angle of projection is
1 \(\mathrm{Y}\)
2 \(2 \mathrm{Y}\)
3 \(3 \mathrm{Y}\)
4 \(\frac{Y}{3}\)
Explanation:
D Given, \(\theta_{1}=\pi / 3, \mathrm{~h}_{1}=\mathrm{Y}, \theta_{2}=\pi / 6\) Let \(h_{2}\) be the maximum height of other stone. We know, \(h=\frac{u^{2} \sin ^{2} \theta}{2 g}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\sin ^{2} \theta_{1}}{\sin ^{2} \theta_{2}}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\sin ^{2}(\pi / 3)}{\sin ^{2}(\pi / 6)}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{(\sqrt{3} / 2)^{2}}{(1 / 2)^{2}}=\frac{3 / 4}{1 / 4}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{3}{1}\) \(\mathrm{h}_{2}=\frac{\mathrm{h}_{1}}{3} \quad\left[\mathrm{~h}_{1}=\mathrm{Y}\right]\) \(\therefore \quad \mathrm{h}_{2}=\frac{\mathrm{Y}}{3}\)
J and K-CET-2000
Motion in Plane
143802
A projectile is thrown with initial velocity \(v_{0}\) and angle \(30^{\circ}\) with the horizontal. If it remains in the air for \(1 \mathrm{sec}\), what was its initial velocity?
1 \(19.6 \mathrm{~m} / \mathrm{s}\)
2 \(9.8 \mathrm{~m} / \mathrm{s}\)
3 \(4.9 \mathrm{~m} / \mathrm{s}\)
4 \(1 \mathrm{~m} / \mathrm{s}\)
Explanation:
B As we know that time of flight in projectile motion is given by \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\) It is given that, \(\mathrm{T}=1 \mathrm{sec}, \quad \theta=30^{\circ}\) \(\mathrm{u}=\mathrm{u}_{0}(\text { initial velocity })\) \(\text { Take } \mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) Time of flight in projectile motion \(1=\frac{2 \mathrm{u}_{0} \sin 30^{\circ}}{9.8} \Rightarrow \mathrm{v}_{0}=\frac{9.8}{2 \times \sin 30^{\circ}}\) \(\mathrm{v}_{0}=9.8 \mathrm{~m} / \mathrm{s}\)
J and K-CET-2013
Motion in Plane
143803
A bird flies at an angle of \(60^{\circ}\) to the horizontal Its horizontal component of velocity is \(10 \mathrm{~m} \mathrm{~s}^{-1}\) . Find the vertical component of velocity in \(\mathrm{m}\) \(\mathbf{s}^{-1}\)
143804
The angle which the velocity vector of a projectile thrown with a velocity \(v\) at an angle \(\theta\) to the horizontal will make with the horizontal after time \(t\) of its being thrown up is:
D As in given question the graph of projectile will be given as Vertical velocity of projectile after time \(\mathrm{t}\) is \(v \sin \theta-\mathrm{gt} \quad\) (from equation of motion) Angle at any time \(t\), \(\tan \phi=\frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}}=\frac{\mathrm{v} \sin \theta-\mathrm{gt}}{\mathrm{v} \cos \theta}\) \(\phi=\tan ^{-1}\left(\frac{\mathrm{v} \sin \theta-\mathrm{gt}}{\mathrm{v} \cos \theta}\right)\)
143799
Two stones are projected with the same magnitude of velocity, but making different angles with horizontal, the angle of projection of one is \(\pi / 3\) and its maximum height is \(Y\), the maximum height attained by the other stone with \(\pi / 6\) angle of projection is
1 \(\mathrm{Y}\)
2 \(2 \mathrm{Y}\)
3 \(3 \mathrm{Y}\)
4 \(\frac{Y}{3}\)
Explanation:
D Given, \(\theta_{1}=\pi / 3, \mathrm{~h}_{1}=\mathrm{Y}, \theta_{2}=\pi / 6\) Let \(h_{2}\) be the maximum height of other stone. We know, \(h=\frac{u^{2} \sin ^{2} \theta}{2 g}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\sin ^{2} \theta_{1}}{\sin ^{2} \theta_{2}}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\sin ^{2}(\pi / 3)}{\sin ^{2}(\pi / 6)}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{(\sqrt{3} / 2)^{2}}{(1 / 2)^{2}}=\frac{3 / 4}{1 / 4}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{3}{1}\) \(\mathrm{h}_{2}=\frac{\mathrm{h}_{1}}{3} \quad\left[\mathrm{~h}_{1}=\mathrm{Y}\right]\) \(\therefore \quad \mathrm{h}_{2}=\frac{\mathrm{Y}}{3}\)
J and K-CET-2000
Motion in Plane
143802
A projectile is thrown with initial velocity \(v_{0}\) and angle \(30^{\circ}\) with the horizontal. If it remains in the air for \(1 \mathrm{sec}\), what was its initial velocity?
1 \(19.6 \mathrm{~m} / \mathrm{s}\)
2 \(9.8 \mathrm{~m} / \mathrm{s}\)
3 \(4.9 \mathrm{~m} / \mathrm{s}\)
4 \(1 \mathrm{~m} / \mathrm{s}\)
Explanation:
B As we know that time of flight in projectile motion is given by \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\) It is given that, \(\mathrm{T}=1 \mathrm{sec}, \quad \theta=30^{\circ}\) \(\mathrm{u}=\mathrm{u}_{0}(\text { initial velocity })\) \(\text { Take } \mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) Time of flight in projectile motion \(1=\frac{2 \mathrm{u}_{0} \sin 30^{\circ}}{9.8} \Rightarrow \mathrm{v}_{0}=\frac{9.8}{2 \times \sin 30^{\circ}}\) \(\mathrm{v}_{0}=9.8 \mathrm{~m} / \mathrm{s}\)
J and K-CET-2013
Motion in Plane
143803
A bird flies at an angle of \(60^{\circ}\) to the horizontal Its horizontal component of velocity is \(10 \mathrm{~m} \mathrm{~s}^{-1}\) . Find the vertical component of velocity in \(\mathrm{m}\) \(\mathbf{s}^{-1}\)
143804
The angle which the velocity vector of a projectile thrown with a velocity \(v\) at an angle \(\theta\) to the horizontal will make with the horizontal after time \(t\) of its being thrown up is:
D As in given question the graph of projectile will be given as Vertical velocity of projectile after time \(\mathrm{t}\) is \(v \sin \theta-\mathrm{gt} \quad\) (from equation of motion) Angle at any time \(t\), \(\tan \phi=\frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}}=\frac{\mathrm{v} \sin \theta-\mathrm{gt}}{\mathrm{v} \cos \theta}\) \(\phi=\tan ^{-1}\left(\frac{\mathrm{v} \sin \theta-\mathrm{gt}}{\mathrm{v} \cos \theta}\right)\)
143799
Two stones are projected with the same magnitude of velocity, but making different angles with horizontal, the angle of projection of one is \(\pi / 3\) and its maximum height is \(Y\), the maximum height attained by the other stone with \(\pi / 6\) angle of projection is
1 \(\mathrm{Y}\)
2 \(2 \mathrm{Y}\)
3 \(3 \mathrm{Y}\)
4 \(\frac{Y}{3}\)
Explanation:
D Given, \(\theta_{1}=\pi / 3, \mathrm{~h}_{1}=\mathrm{Y}, \theta_{2}=\pi / 6\) Let \(h_{2}\) be the maximum height of other stone. We know, \(h=\frac{u^{2} \sin ^{2} \theta}{2 g}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\sin ^{2} \theta_{1}}{\sin ^{2} \theta_{2}}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{\sin ^{2}(\pi / 3)}{\sin ^{2}(\pi / 6)}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{(\sqrt{3} / 2)^{2}}{(1 / 2)^{2}}=\frac{3 / 4}{1 / 4}\) \(\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{3}{1}\) \(\mathrm{h}_{2}=\frac{\mathrm{h}_{1}}{3} \quad\left[\mathrm{~h}_{1}=\mathrm{Y}\right]\) \(\therefore \quad \mathrm{h}_{2}=\frac{\mathrm{Y}}{3}\)
J and K-CET-2000
Motion in Plane
143802
A projectile is thrown with initial velocity \(v_{0}\) and angle \(30^{\circ}\) with the horizontal. If it remains in the air for \(1 \mathrm{sec}\), what was its initial velocity?
1 \(19.6 \mathrm{~m} / \mathrm{s}\)
2 \(9.8 \mathrm{~m} / \mathrm{s}\)
3 \(4.9 \mathrm{~m} / \mathrm{s}\)
4 \(1 \mathrm{~m} / \mathrm{s}\)
Explanation:
B As we know that time of flight in projectile motion is given by \(\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\) It is given that, \(\mathrm{T}=1 \mathrm{sec}, \quad \theta=30^{\circ}\) \(\mathrm{u}=\mathrm{u}_{0}(\text { initial velocity })\) \(\text { Take } \mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\) Time of flight in projectile motion \(1=\frac{2 \mathrm{u}_{0} \sin 30^{\circ}}{9.8} \Rightarrow \mathrm{v}_{0}=\frac{9.8}{2 \times \sin 30^{\circ}}\) \(\mathrm{v}_{0}=9.8 \mathrm{~m} / \mathrm{s}\)
J and K-CET-2013
Motion in Plane
143803
A bird flies at an angle of \(60^{\circ}\) to the horizontal Its horizontal component of velocity is \(10 \mathrm{~m} \mathrm{~s}^{-1}\) . Find the vertical component of velocity in \(\mathrm{m}\) \(\mathbf{s}^{-1}\)
143804
The angle which the velocity vector of a projectile thrown with a velocity \(v\) at an angle \(\theta\) to the horizontal will make with the horizontal after time \(t\) of its being thrown up is:
D As in given question the graph of projectile will be given as Vertical velocity of projectile after time \(\mathrm{t}\) is \(v \sin \theta-\mathrm{gt} \quad\) (from equation of motion) Angle at any time \(t\), \(\tan \phi=\frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}}=\frac{\mathrm{v} \sin \theta-\mathrm{gt}}{\mathrm{v} \cos \theta}\) \(\phi=\tan ^{-1}\left(\frac{\mathrm{v} \sin \theta-\mathrm{gt}}{\mathrm{v} \cos \theta}\right)\)
