The speed of a projectile projected from level ground at its maximum height is found to be half of its speed of projection (u). Its maximum height is:

1.  $\frac{3{\mathrm{u}}^2}{8\mathrm{g}}$

2.  $\frac{3{\mathrm{u}}^2}{2\mathrm{g}}$

3.  $\frac{\sqrt{3}{\mathrm{u}}^2}{2\mathrm{g}}$

4.  $\frac{3{\mathrm{u}}^2}{4\mathrm{g}}$

If ${\mathrm{a}}_{\mathrm{r}}<mspace\; width="1em"></mspace>\mathrm{and}<mspace\; width="1em"></mspace>{\mathrm{a}}_{\mathrm{t}}$ represent radial and tangential accelerations, then the motion of particle will be uniformly circular for :

(1) ${\mathrm{a}}_{\mathrm{r}}$ = 0, ${\mathrm{a}}_{\mathrm{t}}$ = 0

(2) ${\mathrm{a}}_{\mathrm{r}}$ = 0, ${\mathrm{a}}_{\mathrm{t}}$ $\ne$ 0

(3) ${\mathrm{a}}_{\mathrm{r}}$ $\ne$ 0, ${\mathrm{a}}_{\mathrm{t}}$ = 0

(4) ${\mathrm{a}}_{\mathrm{r}}$ $\ne$ 0, ${\mathrm{a}}_{\mathrm{t}}$ $\ne$ 0

If a particle is moving in a circular orbit with constant speed, then:

A body is projected horizontally with speed u from a tower of height h. The magnitude of the velocity with which it will collide with the ground will be: [g is the acceleration due to gravity]

1.  $\sqrt{2\mathrm{gh}}$

2.  $\mathrm{u}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\sqrt{2\mathrm{gh}}$

3.  ${\mathrm{u}}^2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\sqrt{2\mathrm{gh}}$

4.  $\sqrt{{\mathrm{u}}^2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2\mathrm{gh}}$

The position vector of a particle \(\overrightarrow r\) as a function of time \(t\) (in seconds) is \(\overrightarrow r=3 t \hat{i}+2t^2\hat j~\text{m}\). The initial acceleration of the particle is:
1. \(2~\text{m/s}^2\)
2. \(3~\text{m/s}^2\)
3. \(4~\text{m/s}^2\)
4. zero

Coordinates of a particle as a function of time \(t\) are \(x= 2t\),
\(y =4t\). It can be inferred that the path of the particle will be:

At a certain instant, a particle moving in the \(xy\text-\)plane has a velocity of \(\vec v=(2\hat{i}+3\hat{j})~\text{m/s}\) and an acceleration of \(\vec a=(-3\hat{i}+2\hat{j})~\text{m/s}^2.\) What is the rate of change of the particle’s speed at that instant?
1. \(\sqrt{13}\) m/s²
2. \(-1\) m/s²
3. \(1\) m/s²
4. zero

A body started moving with an initial velocity of \(4\) m/s along the east and an acceleration \(1\) m/s² along the north. The velocity of the body just after \(4\) s will be?

A projectile thrown from the ground has horizontal range R. If velocity at the highest point is doubled somehow, the new range will be:

(1) 3 R

(2) 1.5 R

(3) $\sqrt{3}$ R

(4) 2 R