An airplane is moving with a velocity \(u.\) It drops a packet from a height \(h.\) The time \(t\) taken by the packet to reach the ground will be:
1. \( \sqrt{\frac{2 g}{h}} \)
2. \( \sqrt{\frac{2 u}{g}} \)
3. \( \sqrt{\frac{h}{2 g}} \)
4. \( \sqrt{\frac{2 h}{g}}\)

A ball P is dropped vertically and another ball Q is thrown horizontally from the  same height and at the same time. If air resistance is neglected, then 

1. Ball P reaches the ground first

2. Ball Q reaches the ground first

3. Both reach the ground at the same time

4. The respective masses of the two balls will decide the time

The time taken by a block of wood (initially at rest) to slide down a smooth inclined plane \(9.8~\text{m}\) long (angle of inclination is $\mathrm{}$\(30^{\circ}\)) is:

A body sliding on a smooth inclined plane requires \(4\) seconds to reach the bottom starting from the rest at the top. How much time does it take to cover one-fourth distance starting from the rest at the top? 

A bullet is fired with a speed of $1000m/\sec$ in order to hit a target 100 m away. If $g=10m/s^2$, the gun should be aimed:

1. Directly towards the target

2. 5 cm above the target

3. 10 cm above the target

4. 15 cm above the target

A train is moving towards the east and a car is along the north at the same speed. The observed direction of the car to the passenger on the train is:

An aeroplane is moving with horizontal velocity u at height h. The velocity of a packet dropped from it on the earth's surface will be (g is acceleration due to gravity) 

1. $\sqrt{u^2+2gh}$

2. $\sqrt{2gh}$

3. 2 gh

4. $\sqrt{u^2-2gh}$ 

A body is slipping from an inclined plane of height \(h\) and length \(l\). If the angle of inclination is \(\theta\), the time taken by the body to come from the top to the bottom of this inclined plane is:
1. \(\sqrt{\frac{2 h}{g}}\)
2. \(\sqrt{\frac{2 l}{g}}\)
3. \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)
4. \(\sin \theta \sqrt{\frac{2 h}{g}}\)

A frictionless wire \(AB\) is fixed on a sphere of radius \(R\). A very small spherical ball slips on this wire. The time taken by this ball to slip from \(A\) to \(B\) is:
          
1. \(\frac{2 \sqrt{g R}}{g \cos \theta}\)
2. \(2 \sqrt{g R} . \frac{\cos \theta}{g}\)
3. \(2 \sqrt{\frac{R}{g}}\)
4. \(\frac{g R}{\sqrt{g\cos \theta}}\)