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 t² metre. The angle of projection is:

1. ${\sin }^{-1}\left(\frac{{\displaystyle 4}}{{\displaystyle 5}}\right)$

2. ${\sin }^{-1}\left(\frac{{\displaystyle 3}}{{\displaystyle 5}}\right)$

3. ${\sin }^{-1}\left(\frac{{\displaystyle 4}}{{\displaystyle 3}}\right)$

4. ${\sin }^{-1}\left(\frac{{\displaystyle 3}}{{\displaystyle 4}}\right)$

For a given velocity, a projectile has the same range of R for two angles of projection. If t₁ and t₂ are the times of flight in the two cases then:

(1) $t_1{t}_2\propto R^2$

(2) $t_1{t}_2\propto R$

(3) $t_1{t}_2\propto \frac{1}{R}$

(4) $t_1{t}_2\propto \frac{1}{R^2}$

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

(1) $\sqrt{{\left(v\cos \theta \right)}^2+{\left(v\sin \theta \right)}^2}$

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

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

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

A ball is thrown from a point with a speed v₀ at an angle of projection θ. From the same point and at the same instant a person starts running with a constant speed v₀/2 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°

A stone is thrown at an angle θ with the horizontal reaches a maximum height of H. Then the time of flight of stone will be:

(1) $\sqrt{\frac{2H}{g}}$

(2) $2\sqrt{\frac{2H}{g}}$

(3) $\frac{2\sqrt{2H\sin \theta }}{g}$

(4) $\frac{\sqrt{2H\sin \theta }}{g}$

The maximum horizontal range of a projectile is \(400~\text{m}\). The maximum value of height(ever possible) attained by it will be: 
1. \(100~\text{m}\)
2. \(200~\text{m}\)
3. \(400~\text{m}\)
4. \(800~\text{m}\)

A particle is moving eastwards with velocity of 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is- 

1. Zero

2. $\frac{1}{\sqrt{2}}m\text{/}{s}^{\text{2}}$ toward north-west

3. $\frac{1}{\sqrt{2}}m\text{/}{s}^{\text{2}}$ toward north-east

4. $\frac{1}{2}m\text{/}{s}^{\text{2}}$ toward north-west