A projectile is fired from the surface of the earth with a velocity of \(5\) ms^–1 and at an angle \(\theta\) with the horizontal. Another projectile fired from another planet with a velocity of \(3\) ms^–1 at the same angle follows a trajectory that is identical to the trajectory of the projectile fired from the Earth. The value of the acceleration due to gravity on the other planet is: (given \(g=9.8\) ms^–2)
1. \(3.5\) m/s²
2. \(5.9\) m/s²
3. \(16.3\) m/s²
4. \(110.8\) m/s²

The velocity of a projectile at the initial point \(A\) is \(2\hat i+3\hat j~\)m/s. Its velocity (in m/s) at point \(B\) is:
              

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is:
1. $\theta ={\tan }^{-1}\left(\frac{1}{4}\right)$
2. $\theta ={\tan }^{-1}\left(4\right)$
3. $\theta ={\tan }^{-1}\left(2\right)$
4. $\theta ={45}^0$

A missile is fired for a maximum range with an initial velocity of 20 m/s. If g= 10 m/s², then the range of the missile will be:
1. 50 m
2. 60 m
3. 20 m
4. 40 m

A projectile is fired at an angle of \(45^\circ\) with the horizontal. The elevation angle \(\alpha\) of the projectile at its highest point, as seen from the point of projection is:
1. \(60^\circ\)
2. \(tan^{-1}\left ( \frac{1}{2} \right )\)
3. \(tan^{-1}\left ( \frac{\sqrt{3}}{2} \right )\)
4. \(45^\circ\)

The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is:

1. 15º

2. 30º

3. 45º

4. 60º

A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on level ground, the magnitude of change in its momentum will be:

1.  $2\mathrm{mv}$

2.  $\mathrm{mv}/\sqrt{2}$

3.  $\mathrm{mv}\sqrt{2}$

4.  zero

For a projectile projected at angles (45°-θ) and (45°+θ), the horizontal ranges described by the projectile are in the ratio of:
1. 1:1
2. 2:3
3. 1:2
4. 2:1