A body dropped from a height ‘h’ with an initial speed zero,strikes the ground with a velocity 3 km/hour. Another body of same mass dropped from the same height ‘h’ with an initial speed u’ = 4 km/hour. Find the final velocity of second mass, with which its trikes the ground

3 km/hr
4 km/hr
5 km/hr
6 km/hr

Solution

From third equation of motion, v²= u² + 2 as where v & u are final & initial velocity, a is acceleration,s is distance.

For first case v₁= 3km/hour,u₁= 0, a₁=g & s₁=?

A point traversed half of the distance with a velocity v₀. The half of remaining part of the distance was covered with velocity v₁ & second half of remaining part by v₂ velocity.The mean velocity of the point, averaged over the whole time of motion is

\(\frac{v_{0}+v_{1}+v_{2}}{3}\)
\(\frac{v_{0}+v_{1}+v_{2}}{3}\)
\(\frac{2v_{0}+v_{1}+v_{2}}{3}\)
\(\frac{2v_{0}(v_{1}+v_{2})}{(2v_{0}+v_{1}+v_{2})}\)

Solution

Let the total distance be d. Then for first half distance,

Now average speed

A stone is dropped into a well in which the level of water is h below the top of the well. If v is velocity of sound, the time T after which the splash is heard is given by

T= 2h/v
\(T=\sqrt{\left ( \frac{2h}{g} \right )}+\frac{h}{v}\)
\(T=\sqrt{\left ( \frac{2h}{v} \right )}+\frac{h}{g}\)
\(T=\sqrt{\left ( \frac{2h}{g} \right )}+\frac{h}{v}\)

Solution

Time taken by the stone to reach the water level

Figure shows the v-t graph for two particles P and Q. Which of the following statements regarding their relative motion is true ?Their relative velocity is

is zero
is non-zero but constan
continuously decreases
continuously increases

Solution

The difference in velocities is increasing with time as both of them have more constant but different acceleration.

A rubber ball is dropped from a height of 5 metre on a plane where the acceleration due to gravity is same as that on to the surface of the earth. On bouncing, it rises to a height of 1.8 m. On bouncing, the ball loses its velocity by a factor of

³⁄₅
\(\frac{9}{25}\)
²⁄₅
\(\frac{16}{25}\)

Solution

Downward motion

The displacement x of a particle along a straight line at time t is given by : x =\(a_{0}+\frac{a_{1}t}{2}+\frac{a_{2}}{3}t^{2}\). The acceleration of the particle is

\(\frac{a_{2}}{3}\)
\(\frac{a_{2}}{3}\)
\(\frac{a_{2}}{3}\)
\(a_{0}+\frac{a_{2}}{3}\)

Solution

Differentiated twice.

It is given that t = px²+ qx, where x is displacement and t is time. The acceleration of particle at origin is

\(-\frac{2p}{q^{3}}\)
\(-\frac{3p}{q^{3h}}\)
\(-\frac{8p}{q^{3}}\)
\(-\frac{8p}{q^{3}}\)

Solution

Differentiate two times and put x = 0.

The distance through which a body falls in the nth second is h. The distance through which it falls in the next second is

h
h+^g⁄₂
h – g
h + g

Solution

A particle travels half the distance with a velocity of 6 1ms^-1.The remaining half distance is covered with a velocity of 41ms^-1 for half the time and with a velocity of 81ms^-1 for the rest of the half time. What is the velocity of the particle averaged over the whole time of motion ?

9ms^-1
6 ms^-1
5.35 ms^-1
5 ms^-1

Solution

Average velocity for the second half of the distance is

\(=\frac{V_{1}+V_{2}}{2}=\frac{4+8}{2}\)=6 m s^-1

Given that first half distance is covered with a velocity of 6 ms^-1. Therefore, the average velocity for the whole time of motion is 6 ms^-1

A rocket is fired upward from the earth’s surface such that it creates an acceleration of 19.6 2ms^-2. If after 5 s, its engine is switched off, the maximum height of the rocket from earth’s surface would be

980 m
735 m
490 m
245 m

Solution

Velocity when the engine is switched off

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