A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time (Given $g=9.8m/s^2)$

(1) At least 0.8 m/s

(2) Any speed less than 19.6 m/s

(3) Only with speed 19.6 m/s

(4) More than 19.6 m/s

Read the assertion and reason carefully to mark the correct option out of the options given below:

(1) If both assertion and reason are true and the reason is the correct explanation of the assertion.

(2) If both assertion and reason are true but reason is not the correct explanation of the assertion.

(3) If assertion is true but reason is false.

(4) If the assertion and reason both are false.

Assertion : The equation of motion can be applied only if acceleration is along the direction of velocity and is constant.

Reason : If the acceleration of a body is constant then its motion is known as uniform motion.

The position \(x\) of a particle moving along the \(x\)-axis varies with time \(t\) as \(x=20t-5t^2,\) where \(x\) is in meters and \(t\) is in seconds. The particle reverses its direction of motion at:
1. \(x=40~\text{m}\)
2. \(x=10~\text{m}\)
3. \(x=20~\text{m}\)
4. \(x=30~\text{m}\)

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

A body starting from rest moves with uniform acceleration on a horizontal surface. The body covers \(3\) consecutive equal distances from the beginning in time \(t_1, t_2,\text{and}~t_3\)${\mathrm{}}_{}$ seconds. The ratio of \(t_1:t_2:t_3\) is:
1. \(1:2:3\)
2. \(1:\sqrt{2}:\sqrt{3}\)
3. \(1:\left(\sqrt{2}-1\right):\left(\sqrt{3}-\sqrt{2}\right)\)
4. \(\sqrt{3}:\sqrt{2}:1\)$\mathrm{}$

A man starts from point \(P\) and goes \(20\) m due East, then \(5\) m due North, then \(35\) m due West, and finally \(25\) m due South. His displacement from point \(P\) is:

An aircraft is flying horizontally at a height of 1 km from the ground with a speed of 200 m/s. An anti-craft missile launcher kept at a point on the ground which is in the vertical plane of the motion of aircraft. The muzzle velocity of the missile is 600 m/s. The missile is fired at a time when the aircraft was vertically above the anti-craft missile launcher. At what angle with horizontal should it be fired so as to hit the aircraft?

(1) ${\sin }^{-1}\left(\frac{2}{6}\right)$

(2) ${\cos }^{-1}\left(\frac{1}{3}\right)$

(3) ${\tan }^{-1}\left(3\right)$

(4) Just vertically upward

A man can throw a stone to a maximum height 'h'. The greatest horizontal distance up to which he can throw the stone is

(1) h

(2) 2h

(3) $\frac{\mathrm{h}}{2}$

(4) 4h

An oblique projectile is projected with a speed u. It takes time t, to reach the maximum height and time t, to come back to the ground. Air resistance is not neglected, then $\frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}$ is

(1) > 1

(2) < 1

(3) = 1

(4) Depends on the angle of projection

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