143779 Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is \(\pi / 3\) and the maximum height reached by it is \(102 \mathrm{~m}\). Then the maximum height reached by the other in \(\mathrm{m}\) is :

143780 A projectile has initially the same horizontal velocity as it would acquire, if it had moved from rest with uniform acceleration of \(3 \mathrm{~ms}^{-2}\) for 0.5 minute. If the maximum height reached by it is \(80 \mathrm{~m}\), then the angle of projection is: \((\mathrm{g}=\) \(10 \mathrm{~ms}^{-2}\) )

143781 It is possible to project a particle with a given speed in two possible ways so that, it has the same horizontal range \(R\). The product of the times taken by it in two possible ways is :
[g= acceleration due to gravity]

143782 A stone is projected with a velocity \(u\) at angle \(\theta\) with the horizontal reaches maximum height \(H_{1}\), when it is projected with a velocity \(u\) at an angle \(\left(\frac{\pi}{2}-\theta\right)\) with the horizontal, it reaches to a maximum height \(\mathrm{H}_{2}\). The relation between the horizontal range \(r\) of the projectile, \(H_{1}\) and \(\mathrm{H}_{2}\) is:

143783 For a projectile the ratio of maximum height reached to the square of flight time is:
\(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\)

143779 Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is \(\pi / 3\) and the maximum height reached by it is \(102 \mathrm{~m}\). Then the maximum height reached by the other in \(\mathrm{m}\) is :

143780 A projectile has initially the same horizontal velocity as it would acquire, if it had moved from rest with uniform acceleration of \(3 \mathrm{~ms}^{-2}\) for 0.5 minute. If the maximum height reached by it is \(80 \mathrm{~m}\), then the angle of projection is: \((\mathrm{g}=\) \(10 \mathrm{~ms}^{-2}\) )

143781 It is possible to project a particle with a given speed in two possible ways so that, it has the same horizontal range \(R\). The product of the times taken by it in two possible ways is :
[g= acceleration due to gravity]

143782 A stone is projected with a velocity \(u\) at angle \(\theta\) with the horizontal reaches maximum height \(H_{1}\), when it is projected with a velocity \(u\) at an angle \(\left(\frac{\pi}{2}-\theta\right)\) with the horizontal, it reaches to a maximum height \(\mathrm{H}_{2}\). The relation between the horizontal range \(r\) of the projectile, \(H_{1}\) and \(\mathrm{H}_{2}\) is:

143783 For a projectile the ratio of maximum height reached to the square of flight time is:
\(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\)

143779 Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is \(\pi / 3\) and the maximum height reached by it is \(102 \mathrm{~m}\). Then the maximum height reached by the other in \(\mathrm{m}\) is :

143780 A projectile has initially the same horizontal velocity as it would acquire, if it had moved from rest with uniform acceleration of \(3 \mathrm{~ms}^{-2}\) for 0.5 minute. If the maximum height reached by it is \(80 \mathrm{~m}\), then the angle of projection is: \((\mathrm{g}=\) \(10 \mathrm{~ms}^{-2}\) )

143781 It is possible to project a particle with a given speed in two possible ways so that, it has the same horizontal range \(R\). The product of the times taken by it in two possible ways is :
[g= acceleration due to gravity]

143782 A stone is projected with a velocity \(u\) at angle \(\theta\) with the horizontal reaches maximum height \(H_{1}\), when it is projected with a velocity \(u\) at an angle \(\left(\frac{\pi}{2}-\theta\right)\) with the horizontal, it reaches to a maximum height \(\mathrm{H}_{2}\). The relation between the horizontal range \(r\) of the projectile, \(H_{1}\) and \(\mathrm{H}_{2}\) is:

143783 For a projectile the ratio of maximum height reached to the square of flight time is:
\(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\)

143779 Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is \(\pi / 3\) and the maximum height reached by it is \(102 \mathrm{~m}\). Then the maximum height reached by the other in \(\mathrm{m}\) is :

143780 A projectile has initially the same horizontal velocity as it would acquire, if it had moved from rest with uniform acceleration of \(3 \mathrm{~ms}^{-2}\) for 0.5 minute. If the maximum height reached by it is \(80 \mathrm{~m}\), then the angle of projection is: \((\mathrm{g}=\) \(10 \mathrm{~ms}^{-2}\) )

143781 It is possible to project a particle with a given speed in two possible ways so that, it has the same horizontal range \(R\). The product of the times taken by it in two possible ways is :
[g= acceleration due to gravity]

143782 A stone is projected with a velocity \(u\) at angle \(\theta\) with the horizontal reaches maximum height \(H_{1}\), when it is projected with a velocity \(u\) at an angle \(\left(\frac{\pi}{2}-\theta\right)\) with the horizontal, it reaches to a maximum height \(\mathrm{H}_{2}\). The relation between the horizontal range \(r\) of the projectile, \(H_{1}\) and \(\mathrm{H}_{2}\) is:

143783 For a projectile the ratio of maximum height reached to the square of flight time is:
\(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\)

143779 Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is \(\pi / 3\) and the maximum height reached by it is \(102 \mathrm{~m}\). Then the maximum height reached by the other in \(\mathrm{m}\) is :

143780 A projectile has initially the same horizontal velocity as it would acquire, if it had moved from rest with uniform acceleration of \(3 \mathrm{~ms}^{-2}\) for 0.5 minute. If the maximum height reached by it is \(80 \mathrm{~m}\), then the angle of projection is: \((\mathrm{g}=\) \(10 \mathrm{~ms}^{-2}\) )

143781 It is possible to project a particle with a given speed in two possible ways so that, it has the same horizontal range \(R\). The product of the times taken by it in two possible ways is :
[g= acceleration due to gravity]

143782 A stone is projected with a velocity \(u\) at angle \(\theta\) with the horizontal reaches maximum height \(H_{1}\), when it is projected with a velocity \(u\) at an angle \(\left(\frac{\pi}{2}-\theta\right)\) with the horizontal, it reaches to a maximum height \(\mathrm{H}_{2}\). The relation between the horizontal range \(r\) of the projectile, \(H_{1}\) and \(\mathrm{H}_{2}\) is:

143783 For a projectile the ratio of maximum height reached to the square of flight time is:
\(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\)