A ball is falling freely strikes to ground with velocity
60 m/s. Height fallen in last one second before
hitting the ground is

1. 55 m

2. 70 m

3. 60 m

4. 80 m

A bus moves over a straight level road with a constant acceleration a. A body in the bus drops a ball outside. The acceleration of the ball with respect to the bus and the earth are respectively

(1) a and g

(2) a + g and g - a

(3) $\sqrt{a^2+g^2}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>g$

(4) $\sqrt{a^2+g^2}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>a$

A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 s, the average acceleration during contact is nearly (Take g = 10 m$s^{-2}$)

(1) $1500\sqrt{2}<mspace\; width="1em"></mspace>m{s}^{-2}<mspace\; width="1em"></mspace>upwards$

(2) $1800<mspace\; width="1em"></mspace>m{s}^{-2}<mspace\; width="1em"></mspace>downwards$

(3) $1500\sqrt{5}<mspace\; width="1em"></mspace>m{s}^{-2}<mspace\; width="1em"></mspace>upwards$

(4) $1500\sqrt{2}<mspace\; width="1em"></mspace>m{s}^{-2}<mspace\; width="1em"></mspace>downwards$

Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings (building being \(100~\text{m}\) apart and being of the same height of \(200~\text{m}\)) with the same velocity of \(25~\text{m/s}.\) When and where will the two bullets collide?
\((g = 10~\text{m/s}^2)\)

A point moves in a straight line so that its displacement x metre at time t sec is given by ${\mathrm{x}}^2=1+{\mathrm{t}}^2$ . Its acceleration in m/s² at time t sec is

1. $\frac{1}{{\mathrm{x}}^3}$

2. $\frac{1}{\mathrm{x}}-\frac{1}{x^2}$

3. $\frac{1}{\mathrm{x}}-\frac{t^2}{x^3}$

4. $\frac{-\mathrm{t}}{{\mathrm{x}}^2}$

Two particles \(\mathrm{A}\) and \(\mathrm{B}\), move with constant velocities \(\overrightarrow{{v}_1}\) and \(\overrightarrow{{v}_2}\) respectively. At the initial moment, their position vectors are \(\overrightarrow{{r}_1}\) and \(\overrightarrow{{r}_2}\) respectively. The condition for particles \(\mathrm{A}\) and \(\mathrm{B}\) for their collision will be:
1. \(\vec{r_1} \cdot \vec{v_1}=\vec{r_2} \cdot \vec{v_2}\)
2. ​​​​​\(\dfrac{\vec{r_1}-\vec{r_2}}{\left|\vec{r_1}-\vec{r_2}\right|}=\dfrac{\vec{v_2}-\vec{v_1}}{\left|\vec{v_2}-\vec{v_1}\right|}\)${\mathrm{}}_{}$
3. \(\vec{r_1} \times \vec{v_1}=\vec{r_2} \times \vec{v_2}\)${\mathrm{<mspace\; width="1em"></mspace>\; <mspace\; width="1em"></mspace>\; <mspace\; width="1em"></mspace>}}_{}$
4. \(\vec{r_1}-\vec{r_2}=\vec{v_1}-\vec{v_2}\) 

Four persons K,L,M and N are initially at the corners of a square of side of length d. If every person starts moving, such that K is always headed towards L,L towards M,M is headed directly towards N and N towards K, then the four persons will meet after 

1. $\frac{\mathrm{d}}{\mathrm{v}}<mspace\; width="1em"></mspace>sec$

2. $\frac{\sqrt{2\mathrm{d}}}{\mathrm{v}}<mspace\; width="1em"></mspace>sec$

3. $\frac{d}{\sqrt{2\mathrm{v}}}<mspace\; width="1em"></mspace>sec$

4. $\frac{\mathrm{d}}{2\mathrm{v}}<mspace\; width="1em"></mspace>sec$

The coordinates of a moving particle at a time t, are given by, x=5 sin 10t, y=5 cos 10t. The speed of the particle is:

(A)  25

(B)  50

(C)  10

(D)  $50\sqrt{2}$

The engine of a motorcycle can produce a maximum acceleration 5 m/s². Its brakes can produce a maximum retardation 10 m/s². What is the minimum time in which it can cover a distance of 1.5 km?

(1) 30 sec

(2) 15 sec

(3) 10 sec

(4) 5 sec