In a projectile motion, velocity at maximum height is 

(1) $\frac{u\cos \theta }{2}$

(2) $u\cos \theta$

(3) $\frac{u\sin \theta }{2}$

(4) None of these

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 cricketer can throw a ball to a maximum horizontal distance of \(100~\text{m}\). With the same effort, he throws the ball vertically upwards. The maximum height attained by the ball is: 
1. \(100~\text{m}\)
2. \(80~\text{m}\)
3. \(60~\text{m}\)
4. \(50~\text{m}\)

A ball is projected with velocity v₀ at an angle of elevation 30°. Mark the correct statement.

(1) Kinetic energy will be zero at the highest point of the trajectory.

(2) Vertical component of momentum will be conserved.

(3) Horizontal component of momentum will be conserved.

(4) Gravitational potential energy will be minimum at the highest point of the trajectory.

Neglecting the air resistance, the time of flight of a projectile is determined by:

(1) U_vertical

(2) U_horizontal

(3) U = U²_vertical + U²_horizontal

(4) U = U(U²_vertical + U²_horizontal )^1/2

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 horizontal range of a projectile is \(4 \sqrt{3}\) times its maximum height. Its angle of projection will be:
1. \(45^{\circ}\)
2. \(60^{\circ}\)
3. \(90^{\circ}\)
4. \(30^{\circ}\)

The maximum horizontal range of a projectile is 400 m. The maximum value of height(ever possible) attained by it will be 

(1) 100 m

(2) 200 m

(3) 400 m

(4) 800 m