An aeroplane is flying horizontally with a velocity \( u = 600\ \text{km/h}\) at a height of \(1960\ \text{m}\). When it is vertically at a point \(A\) on the ground, a bomb is released from it. The bomb strikes the ground at point \(B\). The distance \(AB\) is:

1. \(1200\ \text{m}\)

2. \(0.33\ \text{km}\)

3. \(3.33\ \text{km}\)

4.  \(33\ \text{km}\)

A particle \((A)\) is dropped from a height and another particle \((B)\) is thrown in the horizontal direction with a speed of \(5\ \text{m/s}\) from the same height. The correct statement is:

1. Both particles will reach at ground simultaneously

2. Both particles will reach at ground with same speed

3. Particle (A) will reach at ground first with respect to particle \((B)\)

4. Particle (B) will reach at ground first with respect to particle \((A)\)

An object is thrown along a direction inclined at an angle of \(45^°\) with the horizontal direction. The horizontal range of the particle is equal to:

1. Vertical height

2. Twice the vertical height

3. Thrice the vertical height

4. Four times the vertical height

If the range of a gun that fires a shell with muzzle speed v is R, then the angle of elevation of the gun is 

(1) ${\cos }^{-1}\left(\frac{{\displaystyle v^2}}{{\displaystyle Rg}}\right)$

(2) ${\cos }^{-1}\left(\frac{{\displaystyle gR}}{{\displaystyle v^2}}\right)$

(3) $\frac{1}{2}\left(\frac{{\displaystyle v^2}}{{\displaystyle Rg}}\right)$

(4) $\frac{1}{2}{\sin }^{-1}\left(\frac{{\displaystyle gR}}{{\displaystyle v^2}}\right)$

Four bodies \(P\), \(Q\), \(R\) and \(S\) are projected with equal velocities having angles of projection \(15^{\circ},\) \(30^{\circ},\)\(45^{\circ},\) and \(60^{\circ}\) with the horizontal respectively. The body having the shortest range is? 

Which of the following sets of factors will affect the horizontal distance covered by an athlete in a long–jump event?
1. Speed before he jumps and his weight
2. The direction in which he leaps and the initial speed
3. The force with which he pushes the ground and his speed
4. None of the above

The equation of motion of a projectile is given by \(x = 36 t\) metre and \(2y = 96 t – 9.8 t^2\) metre. The angle of projection is:

1. \(\sin^{− 1} \frac 45\)

2. \(\sin^{− 1} \frac 35\)

3. \(\sin^{− 1} \frac 43\)

4. \(\sin^{− 1} \frac 34\)${}^{}$

For a given velocity, a projectile has the same range of \(R\) for two angles of projection. If \(t_1\) and \(t_2\) are the times of flight in the two cases then:
1. \(t_1t_2 \propto R^2\)

2. \(t_1t_2 \propto R\)

3. \(t_1t_2 \propto \dfrac{1}{R}\)

4. \(t_1t_2 \propto \dfrac{1}{R^2}\)

A body of mass \(m\) is thrown upwards at an angle \(θ\) with the horizontal at velocity \(v\). While rising, the velocity of the mass after \(t\) seconds will be:

1. \(\sqrt{(v\cos\theta)^2+(v\sin\theta)^2}\)

2. \(\sqrt{(v\cos\theta-v\sin\theta)^2-gt}\)

3. \(\sqrt{v^2+g^2t^2-(2v\sin\theta)gt}\)

4. \(\sqrt{v^2+g^2t^2-(2v\cos\theta)gt}\)

A ball is thrown from a point with a speed \(v_0\) at an angle of projection \(θ\). From the same point and at the same instant, a person starts running with a constant speed \(v_0/2\) to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?

1. Yes, \(60^°\)

2. Yes, \(30^°\)

3. No

4. Yes, \(45^°\)