The position of a particle at time \(t\) is given by, \(x=3t^3\), \(y=2t^2+8t\), and \(z=6t-5\). The initial velocity of the particle is:

The equation of the trajectory of a projectile is given as $\mathrm{y}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>4\mathrm{x}<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>\frac{1}{4}{\mathrm{x}}^2$. The range of the projectile is:

1. 1 m

2. 32 m

3. 16 m

4. 48 m

A projectile is fired from the surface of the earth with a velocity of 5ms^–1 and at an angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3ms^–1 at the same angle follows a trajectory that is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is: (given = 9.8 ms^–2)

1. 3.5

2. 5.9

3. 16.3

4. 110.8

The horizontal range is four times the maximum height attained by a projectile. The angle of projection is

1.  90$°$

2.  60$°$

3.  45$°$

4.  30$°$

Two bullets are fired horizontally with different velocities from the same height. Which will reach the ground first?

Two projectiles are projected at angles $\left(\frac{\mathrm{\pi }}{4}+\mathrm{\theta }\right)$ and $\left(\frac{\mathrm{\pi }}{4}-\mathrm{\theta }\right)$ with the horizontal, where $\mathrm{\theta }<\frac{\mathrm{\pi }}{4}$, with same speed. The ratio of horizontal ranges described by them is

1. $\tan <mspace\; width="1em"></mspace>\mathrm{\theta }<mspace\; width="1em"></mspace>:<mspace\; width="1em"></mspace>1$

2. $1<mspace\; width="1em"></mspace>:<mspace\; width="1em"></mspace>{\tan }^2<mspace\; width="1em"></mspace>\mathrm{\theta }$

3. $1<mspace\; width="1em"></mspace>:<mspace\; width="1em"></mspace>1$

4. $1<mspace\; width="1em"></mspace>:<mspace\; width="1em"></mspace>\sqrt{3}$