Let particle have charge q and mass 'm

We know, F = ma_y = qE

$$ \therefore $$ $${a_y} = {{qE} \over m}$$

As electric field is along vertical direction so horizontal component of acceleration is zero.

$$ \therefore $$ a_x = 0

Applying, $$S = {u_y}t + {1 \over 2}{a_y}{t^2}$$

$$ - y = 0\times t + {1 \over 2}\left( {{{qE} \over m}} \right){t^2}$$ ....(1)

Also, $$x = {V_0}t$$

$$ \Rightarrow t = {x \over {{V_0}}}$$

Putting in (1) we get,

$$y = - {1 \over 2}\left( {{{qE} \over m}} \right){\left( {{x \over {{V_0}}}} \right)^2}$$

Differentiating with respect to x,

$${{dy} \over {dx}} = - {{qE} \over m}\left( {{x \over {V_0^2}}} \right)$$

$$ \therefore $$ Slope = $$m = \tan \theta = - {{qE} \over m}\left( {{x \over {V_0^2}}} \right)$$

when x = d, then $$m = - {{qE} \over m}\left( {{d \over {V_0^2}}} \right)$$

We know, equation of straight line whose slope is m,

y = mx + c

Here m = $$\left( { - {{qEd} \over {mV_0^2}}} \right)$$,

$$ \therefore $$ $$ - {1 \over 2}\left( {{{qE} \over m}} \right){\left( {{d \over {{V_0}}}} \right)^2} = \left( { - {{qEd} \over {mV_0^2}}} \right)d + c$$

$$ \Rightarrow $$ c = $${{{qEd} \over {2mV_0^2}}}$$

Putting the value of c, we get

$$y = - \left( {{{qEd} \over {mV_0^2}}} \right)x + {{qE{d^2}} \over {2mV_0^2}}$$

$$ = {{qEd} \over {mV_0^2}}\left( {{d \over 2} - x} \right)$$