362003
A particle is projected with speed \(u\) that makes an angle \(\theta \) with horizontal, the time at which the velocity of the projectile is perpendicular to the initial velocity is
1 \(\frac{u}{{g\cos \theta }}\)
2 \(\frac{u}{{g\tan \theta }}\)
3 \(\frac{u}{{g\sin \theta }}\)
4 \(\frac{{u\sin \theta }}{g}\)
Explanation:
The initial velocity vector is \(\overrightarrow u = u\cos \theta \hat i + u\sin \theta \hat j\) The velocity vector after time \(t\) is \(\overrightarrow v = u\cos \theta \hat i + \left( {u\sin \theta - gt} \right)\hat j\) When \(\overrightarrow v \) is perpendicular to \(\overrightarrow u \) \(\overrightarrow v .\overrightarrow u = 0\) \({u^2}{\cos ^2} + {u^2}{\sin ^2}\theta - u\sin \theta gt = 0\) \(t = \frac{u}{{g\sin \theta }}\)
PHXI04:MOTION IN A PLANE
362004
A projectile of mass \(m\) is thrown with a velocity \(v\) making an angle of \(45^\circ \) with the horizontal. The change in momentum from departure to arrival along vertical direction, is
1 \(2\,mv\)
2 \(\sqrt 2 \,mv\)
3 \(mv\)
4 \(\frac{{4R}}{g}\)
Explanation:
Change in momentum is the product of force and time. \(\Delta p = mg \times \frac{{2\sin \theta }}{g}\) \(\left( {\because \;F = \frac{{\Delta p}}{{\Delta t}}} \right)\) \( = 2mv\sin \theta = 2mv\sin 45^\circ \) \( = \frac{{2mv}}{{\sqrt 2 }} = \sqrt 2 mv\)
NCERT Exemplar
PHXI04:MOTION IN A PLANE
362005
A projectile is fired making an angle \(2\theta \) with horizontal with velocity \(4m{s^{ - 1}}\). At some instant it makes an angle \({\rm{\theta }}\), then its velocity is :
362003
A particle is projected with speed \(u\) that makes an angle \(\theta \) with horizontal, the time at which the velocity of the projectile is perpendicular to the initial velocity is
1 \(\frac{u}{{g\cos \theta }}\)
2 \(\frac{u}{{g\tan \theta }}\)
3 \(\frac{u}{{g\sin \theta }}\)
4 \(\frac{{u\sin \theta }}{g}\)
Explanation:
The initial velocity vector is \(\overrightarrow u = u\cos \theta \hat i + u\sin \theta \hat j\) The velocity vector after time \(t\) is \(\overrightarrow v = u\cos \theta \hat i + \left( {u\sin \theta - gt} \right)\hat j\) When \(\overrightarrow v \) is perpendicular to \(\overrightarrow u \) \(\overrightarrow v .\overrightarrow u = 0\) \({u^2}{\cos ^2} + {u^2}{\sin ^2}\theta - u\sin \theta gt = 0\) \(t = \frac{u}{{g\sin \theta }}\)
PHXI04:MOTION IN A PLANE
362004
A projectile of mass \(m\) is thrown with a velocity \(v\) making an angle of \(45^\circ \) with the horizontal. The change in momentum from departure to arrival along vertical direction, is
1 \(2\,mv\)
2 \(\sqrt 2 \,mv\)
3 \(mv\)
4 \(\frac{{4R}}{g}\)
Explanation:
Change in momentum is the product of force and time. \(\Delta p = mg \times \frac{{2\sin \theta }}{g}\) \(\left( {\because \;F = \frac{{\Delta p}}{{\Delta t}}} \right)\) \( = 2mv\sin \theta = 2mv\sin 45^\circ \) \( = \frac{{2mv}}{{\sqrt 2 }} = \sqrt 2 mv\)
NCERT Exemplar
PHXI04:MOTION IN A PLANE
362005
A projectile is fired making an angle \(2\theta \) with horizontal with velocity \(4m{s^{ - 1}}\). At some instant it makes an angle \({\rm{\theta }}\), then its velocity is :
362003
A particle is projected with speed \(u\) that makes an angle \(\theta \) with horizontal, the time at which the velocity of the projectile is perpendicular to the initial velocity is
1 \(\frac{u}{{g\cos \theta }}\)
2 \(\frac{u}{{g\tan \theta }}\)
3 \(\frac{u}{{g\sin \theta }}\)
4 \(\frac{{u\sin \theta }}{g}\)
Explanation:
The initial velocity vector is \(\overrightarrow u = u\cos \theta \hat i + u\sin \theta \hat j\) The velocity vector after time \(t\) is \(\overrightarrow v = u\cos \theta \hat i + \left( {u\sin \theta - gt} \right)\hat j\) When \(\overrightarrow v \) is perpendicular to \(\overrightarrow u \) \(\overrightarrow v .\overrightarrow u = 0\) \({u^2}{\cos ^2} + {u^2}{\sin ^2}\theta - u\sin \theta gt = 0\) \(t = \frac{u}{{g\sin \theta }}\)
PHXI04:MOTION IN A PLANE
362004
A projectile of mass \(m\) is thrown with a velocity \(v\) making an angle of \(45^\circ \) with the horizontal. The change in momentum from departure to arrival along vertical direction, is
1 \(2\,mv\)
2 \(\sqrt 2 \,mv\)
3 \(mv\)
4 \(\frac{{4R}}{g}\)
Explanation:
Change in momentum is the product of force and time. \(\Delta p = mg \times \frac{{2\sin \theta }}{g}\) \(\left( {\because \;F = \frac{{\Delta p}}{{\Delta t}}} \right)\) \( = 2mv\sin \theta = 2mv\sin 45^\circ \) \( = \frac{{2mv}}{{\sqrt 2 }} = \sqrt 2 mv\)
NCERT Exemplar
PHXI04:MOTION IN A PLANE
362005
A projectile is fired making an angle \(2\theta \) with horizontal with velocity \(4m{s^{ - 1}}\). At some instant it makes an angle \({\rm{\theta }}\), then its velocity is :
362003
A particle is projected with speed \(u\) that makes an angle \(\theta \) with horizontal, the time at which the velocity of the projectile is perpendicular to the initial velocity is
1 \(\frac{u}{{g\cos \theta }}\)
2 \(\frac{u}{{g\tan \theta }}\)
3 \(\frac{u}{{g\sin \theta }}\)
4 \(\frac{{u\sin \theta }}{g}\)
Explanation:
The initial velocity vector is \(\overrightarrow u = u\cos \theta \hat i + u\sin \theta \hat j\) The velocity vector after time \(t\) is \(\overrightarrow v = u\cos \theta \hat i + \left( {u\sin \theta - gt} \right)\hat j\) When \(\overrightarrow v \) is perpendicular to \(\overrightarrow u \) \(\overrightarrow v .\overrightarrow u = 0\) \({u^2}{\cos ^2} + {u^2}{\sin ^2}\theta - u\sin \theta gt = 0\) \(t = \frac{u}{{g\sin \theta }}\)
PHXI04:MOTION IN A PLANE
362004
A projectile of mass \(m\) is thrown with a velocity \(v\) making an angle of \(45^\circ \) with the horizontal. The change in momentum from departure to arrival along vertical direction, is
1 \(2\,mv\)
2 \(\sqrt 2 \,mv\)
3 \(mv\)
4 \(\frac{{4R}}{g}\)
Explanation:
Change in momentum is the product of force and time. \(\Delta p = mg \times \frac{{2\sin \theta }}{g}\) \(\left( {\because \;F = \frac{{\Delta p}}{{\Delta t}}} \right)\) \( = 2mv\sin \theta = 2mv\sin 45^\circ \) \( = \frac{{2mv}}{{\sqrt 2 }} = \sqrt 2 mv\)
NCERT Exemplar
PHXI04:MOTION IN A PLANE
362005
A projectile is fired making an angle \(2\theta \) with horizontal with velocity \(4m{s^{ - 1}}\). At some instant it makes an angle \({\rm{\theta }}\), then its velocity is :
362003
A particle is projected with speed \(u\) that makes an angle \(\theta \) with horizontal, the time at which the velocity of the projectile is perpendicular to the initial velocity is
1 \(\frac{u}{{g\cos \theta }}\)
2 \(\frac{u}{{g\tan \theta }}\)
3 \(\frac{u}{{g\sin \theta }}\)
4 \(\frac{{u\sin \theta }}{g}\)
Explanation:
The initial velocity vector is \(\overrightarrow u = u\cos \theta \hat i + u\sin \theta \hat j\) The velocity vector after time \(t\) is \(\overrightarrow v = u\cos \theta \hat i + \left( {u\sin \theta - gt} \right)\hat j\) When \(\overrightarrow v \) is perpendicular to \(\overrightarrow u \) \(\overrightarrow v .\overrightarrow u = 0\) \({u^2}{\cos ^2} + {u^2}{\sin ^2}\theta - u\sin \theta gt = 0\) \(t = \frac{u}{{g\sin \theta }}\)
PHXI04:MOTION IN A PLANE
362004
A projectile of mass \(m\) is thrown with a velocity \(v\) making an angle of \(45^\circ \) with the horizontal. The change in momentum from departure to arrival along vertical direction, is
1 \(2\,mv\)
2 \(\sqrt 2 \,mv\)
3 \(mv\)
4 \(\frac{{4R}}{g}\)
Explanation:
Change in momentum is the product of force and time. \(\Delta p = mg \times \frac{{2\sin \theta }}{g}\) \(\left( {\because \;F = \frac{{\Delta p}}{{\Delta t}}} \right)\) \( = 2mv\sin \theta = 2mv\sin 45^\circ \) \( = \frac{{2mv}}{{\sqrt 2 }} = \sqrt 2 mv\)
NCERT Exemplar
PHXI04:MOTION IN A PLANE
362005
A projectile is fired making an angle \(2\theta \) with horizontal with velocity \(4m{s^{ - 1}}\). At some instant it makes an angle \({\rm{\theta }}\), then its velocity is :
