A particle is dropped from a tower in a uniform gravitational field at t = 0. The particle is blown over by a horizontal wind with constant velocity. The slope (m) of the trajectory of the particle with horizontal and its kinetic energy vary according to curves. Here, x is the horizontal displacement and h is the height of particle from ground at time t.

Initially, accelerations are opposite to velocities. Hence, motion will be retarded. But after sometimes velocity will become zero and then velocity will in the direction of acceleration. Now the motion will be accelerated. 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 v₀. Then it may be assumed that the particle is projected horizontally from the top of the tower with velocity v₀. 

Hence, for the particle, initial velocity u = v₀ and angle of projection $\mathrm{\theta }$ = 0^o. 

We know equation of trajectory is $\mathrm{y}=\mathrm{xtan}\mathrm{\theta }-\frac{{\mathrm{gx}}^2}{2{\mathrm{u}}^2{\cos }^2\mathrm{\theta }}$

Here, $\mathrm{y}=-\frac{{\mathrm{gx}}^2}{2{\mathrm{v}}_0^2}\left(\text{ putting }\mathrm{\theta }=0^{\circ }\right)$

The slope of the trajectory of the particle is 

$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2\mathrm{gx}}{2{\mathrm{v}}_0^2}=-\frac{\mathrm{g}}{{\mathrm{v}}_0^2}\mathrm{x}$

Hence, the curve between slope and x 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, x = v₀t.

We get $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{gt}}{{\mathrm{v}}_0}$

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

The vertical component of velocity of the particle at time t is equal to gt. Hence, at time t,
$\mathrm{KE}=\frac{1}{2}\mathrm{m}\left[(\mathrm{gt}{)}^2+{\left({\mathrm{v}}_0\right)}^2\right]$

It means, the graph between KE and time t should be a parabola having value $\frac{1}{2}{\mathrm{mv}}_0^2$ at t = 0. Therefore, option (c) is correct.

As the particle falls, its height decreases and KE increases.
$\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}_0^2+\mathrm{mg}(\mathrm{H}-\mathrm{h})$, where H 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.