361991
An object is projected with a kinetic energy \(k\). Its range is \(R\). It will have the minimum kinetic energy after travelling a horizontal distance equal to
1 \(0.25R\)
2 \(0.5R\)
3 \(0.75R\)
4 \(R\)
Explanation:
At the maximum height potential energy is maximum and hence its kinetic energy is minimum at which the body covers half of the range along \(x\)-axis
PHXI04:MOTION IN A PLANE
361992
Velocity of a projectile in its flight
1 Remains constant
2 First decreases, becomes zero and then increases.
3 First decreases, reaches minimum and then increases.
4 First increases, reaches maximum and then decreases.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361993
A projectile is thrown with a velocity of \(10\sqrt 2 \,m{s^{ - 1}}\) at angle \(45^\circ \) with horizontal. The interval between the moments when speed is \(\sqrt {125} \,m{s^{ - 1}}\) is (Take \(g = 10\,m{s^{ - 2}}\))
361994
A particle has been projected at certain angle \(\theta \) with the horizontal. Find the value of \(\theta \) for which the particle hits the ground in such a manner that initial and final velocity vectors are at \(90^\circ \)
1 \(60^\circ \)
2 \(30^\circ \)
3 \(45^\circ \)
4 For any value of \(\theta \)
Explanation:
Let the intial and final velocities of the projectile are \({\overline V _i} = u\cos \theta \widehat i + u\sin \theta \widehat j\) \({\overline V _f} = u\cos \theta \widehat i - u\sin \theta \widehat j\) \(u - \) intial speed Given that \({\overline V _i} \cdot {\overline V _f} = 0 \Rightarrow \theta = {45^0}\)
361991
An object is projected with a kinetic energy \(k\). Its range is \(R\). It will have the minimum kinetic energy after travelling a horizontal distance equal to
1 \(0.25R\)
2 \(0.5R\)
3 \(0.75R\)
4 \(R\)
Explanation:
At the maximum height potential energy is maximum and hence its kinetic energy is minimum at which the body covers half of the range along \(x\)-axis
PHXI04:MOTION IN A PLANE
361992
Velocity of a projectile in its flight
1 Remains constant
2 First decreases, becomes zero and then increases.
3 First decreases, reaches minimum and then increases.
4 First increases, reaches maximum and then decreases.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361993
A projectile is thrown with a velocity of \(10\sqrt 2 \,m{s^{ - 1}}\) at angle \(45^\circ \) with horizontal. The interval between the moments when speed is \(\sqrt {125} \,m{s^{ - 1}}\) is (Take \(g = 10\,m{s^{ - 2}}\))
361994
A particle has been projected at certain angle \(\theta \) with the horizontal. Find the value of \(\theta \) for which the particle hits the ground in such a manner that initial and final velocity vectors are at \(90^\circ \)
1 \(60^\circ \)
2 \(30^\circ \)
3 \(45^\circ \)
4 For any value of \(\theta \)
Explanation:
Let the intial and final velocities of the projectile are \({\overline V _i} = u\cos \theta \widehat i + u\sin \theta \widehat j\) \({\overline V _f} = u\cos \theta \widehat i - u\sin \theta \widehat j\) \(u - \) intial speed Given that \({\overline V _i} \cdot {\overline V _f} = 0 \Rightarrow \theta = {45^0}\)
361991
An object is projected with a kinetic energy \(k\). Its range is \(R\). It will have the minimum kinetic energy after travelling a horizontal distance equal to
1 \(0.25R\)
2 \(0.5R\)
3 \(0.75R\)
4 \(R\)
Explanation:
At the maximum height potential energy is maximum and hence its kinetic energy is minimum at which the body covers half of the range along \(x\)-axis
PHXI04:MOTION IN A PLANE
361992
Velocity of a projectile in its flight
1 Remains constant
2 First decreases, becomes zero and then increases.
3 First decreases, reaches minimum and then increases.
4 First increases, reaches maximum and then decreases.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361993
A projectile is thrown with a velocity of \(10\sqrt 2 \,m{s^{ - 1}}\) at angle \(45^\circ \) with horizontal. The interval between the moments when speed is \(\sqrt {125} \,m{s^{ - 1}}\) is (Take \(g = 10\,m{s^{ - 2}}\))
361994
A particle has been projected at certain angle \(\theta \) with the horizontal. Find the value of \(\theta \) for which the particle hits the ground in such a manner that initial and final velocity vectors are at \(90^\circ \)
1 \(60^\circ \)
2 \(30^\circ \)
3 \(45^\circ \)
4 For any value of \(\theta \)
Explanation:
Let the intial and final velocities of the projectile are \({\overline V _i} = u\cos \theta \widehat i + u\sin \theta \widehat j\) \({\overline V _f} = u\cos \theta \widehat i - u\sin \theta \widehat j\) \(u - \) intial speed Given that \({\overline V _i} \cdot {\overline V _f} = 0 \Rightarrow \theta = {45^0}\)
361991
An object is projected with a kinetic energy \(k\). Its range is \(R\). It will have the minimum kinetic energy after travelling a horizontal distance equal to
1 \(0.25R\)
2 \(0.5R\)
3 \(0.75R\)
4 \(R\)
Explanation:
At the maximum height potential energy is maximum and hence its kinetic energy is minimum at which the body covers half of the range along \(x\)-axis
PHXI04:MOTION IN A PLANE
361992
Velocity of a projectile in its flight
1 Remains constant
2 First decreases, becomes zero and then increases.
3 First decreases, reaches minimum and then increases.
4 First increases, reaches maximum and then decreases.
Explanation:
Conceptual Question
PHXI04:MOTION IN A PLANE
361993
A projectile is thrown with a velocity of \(10\sqrt 2 \,m{s^{ - 1}}\) at angle \(45^\circ \) with horizontal. The interval between the moments when speed is \(\sqrt {125} \,m{s^{ - 1}}\) is (Take \(g = 10\,m{s^{ - 2}}\))
361994
A particle has been projected at certain angle \(\theta \) with the horizontal. Find the value of \(\theta \) for which the particle hits the ground in such a manner that initial and final velocity vectors are at \(90^\circ \)
1 \(60^\circ \)
2 \(30^\circ \)
3 \(45^\circ \)
4 For any value of \(\theta \)
Explanation:
Let the intial and final velocities of the projectile are \({\overline V _i} = u\cos \theta \widehat i + u\sin \theta \widehat j\) \({\overline V _f} = u\cos \theta \widehat i - u\sin \theta \widehat j\) \(u - \) intial speed Given that \({\overline V _i} \cdot {\overline V _f} = 0 \Rightarrow \theta = {45^0}\)
