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Question ID - 97820 | SaraNextGen Top Answer

A particle is dropped from a tower in a uniform gravitational field at . The particle is blown over by a horizontal wind with constant velocity. Slop  of trajectory of the particle with horizontal and its kinetic energy vary according to curves. Here,  is horizontal displacement and  is height of particle from ground at time

a)b)c)d)

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Answer Key / Explanation : (b,c) -

(b,c)

Initially, accelerations are opposite to velocities, hence motion will be retarded. But after sometimes velocity will be retarded. But after sometimes velocity will become zero and then velocity will in the direction of acceleration. Now the motion will be acceleration. As the particle is blown over by a wind with constant velocity along horizontal direction, the particle has a horizontal component of velocity. Let this component be . Then it may be assumed that the particle is projected horizontally from the top of the tower with velocity

Hence, for the particle, initial velocity  and angle of projection

We know equation of trajectory is

Here, (putting )

The slope of the trajectory of the particle is

Hence, the curve between slope and  will be a straight line passing through the origin and will have a negative slope. It means that option (b) is correct

Since horizontal velocity of the particle remains constant, . We get

So the graph between  and time  will have the same shape as the graph between and . Hence, option (a) is wrong

The vertical component of velocity of the particle at time  is equal to . Hence, at time

It means, the graph between KE and time  should be a parabola having value at . Therefore, option (c) is correct

As the particle falls, its height decreases and KE increases

where  is the initial height. The KE increases linearly with height of its fall or the graph between KE and height of the particle will be a straight line having negative slope. Hence, option (d) is wrong


127 votes

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