A ball is thrown upward with a certain speed. It passes through the same point at 3 second and 7 second from the start. The maximum height achieved by the ball is

1.  500 m

2.  250 m

3.  125 m

4.  450 m

For the following acceleration versus time graph the corresponding velocity versus displacement graph is 

1. 

2. 

3. 

4. 

A particle is moving along the path y = ${\mathrm{x}}^2$ from x = 0 m to x = 2 m. Then the distance traveled by the particle is:

1.  4 m

2.  $\sqrt{20} \mathrm{m}$

3.  $>\sqrt{20} \mathrm{m}$

4.  $<\sqrt{20} \mathrm{m}$

Six particles situated at the corners of a regular hexagon of side 'a' move at constant speed v. Each particle maintains a direction towards the particle at the next. The time which the particles will take to meet each other is-

1.  $\frac{2\mathrm{a}}{\mathrm{v}} \sec$

2.  $\frac{\mathrm{a}}{\mathrm{v}} \sec$

3.  $\frac{2\mathrm{a}}{3\mathrm{v}} \sec$

4.  $\frac{3\mathrm{a}}{\mathrm{v}} \sec$

The displacement x of a particle moving in one dimension under the action of the constant force is related to time t by the equation $t = \sqrt{x} + 3$, where x is in meters and t is in seconds. Find the displacement of the particle from t = 0 sec to t = 6 s.

1.  0

2.  12 m

3.  6 m

4.  18 m

The acceleration a (in ${\mathrm{ms}}^{-2}$) of a body, starting from rest varies with time t(in s) following the equation a = 3t+4. The velocity of the body at time t = 2s will be :

1. 10 ${\mathrm{ms}}^{-1}$

2. 18 ${\mathrm{ms}}^{-1}$

3. 14 ${\mathrm{ms}}^{-1}$

4. 26 ${\mathrm{ms}}^{-1}$

A body thrown vertically so as to reach its maximum height in t second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is:

1. $\sqrt{2}t$

2. $\left(1+\frac{1}{\sqrt{2}}\right)t$

3. $\frac{3t}{2}$

4. $\frac{t}{\sqrt{2}}$

A stone falls freely from rest from a height h and it travels a distance $\frac{9h}{25}$ in the last second. The value of h is:

1. 145 m

2. 100 m

3. 122.5 m

4. 200 ms

A point moves in a straight line under the retardation a${\mathrm{v}}^2$. If the initial velocity is u, the distance covered in 't' seconds is-

1.  $\mathrm{aut}$

2.  $\frac{1}{\mathrm{a}}\ln \left(\mathrm{aut}\right)$

3.  $\frac{1}{\mathrm{a}}\ln \left(1+\mathrm{aut}\right)$

4.  $\mathrm{a} \ln \left(\mathrm{aut}\right)$

A particle is thrown upwards from ground. It experiences a constant resistance force which can produce retardation of 2 $m/s^2$. The ratio of time of ascent to the time of descent is: $[g=10 m/s^2]$

(1) 1:1

(2) $\sqrt{\frac{2}{3}}$

(3) $\frac{2}{3}$

(4) $\sqrt{\frac{3}{2}}$