(a)If an electron is projected horizontally with a speed of $\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{m}\mathbf{/}\mathbf{s}$, how far will it fall in traversing$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{m}$of horizontal distance?

(b)Does the answer increase or decrease if the initial speed is increased?

2) The horizontal distance traveled by the electron is,$\mathrm{x}=1.0\mathrm{m}$.

In the projectile motion, when the object is thrown horizontally, there is no acceleration in the horizontal direction. Therefore, the object moves with constant speed in the horizontal direction. However, in the vertical direction, it is acted upon by gravitational acceleration. Therefore, the motion in the vertical direction can be described using the kinematic equations. Using kinematic equations and treating the electron as a projectile, we can solve the given problem for the required quantities.

Formula:

$V_f=V_i+at$

Where $V_f$is the final velocity, $V_i$is the initial velocity, $a$is the acceleration, and t is the time.

$∆y=V_{i}t+\frac{1}{2}a{t}^2$

To determine the vertical fall, we first need to calculate the time required to move horizontally through a distance of $1.0\mathrm{m}$

We use the definition of speed to calculate the time since horizontal motion is not an accelerated motion.

i.e.$a_x=0$

Hence,

$\begin{array}{l}{v}_x=\frac{x}{t}\\ t=\frac{x}{v_x}\end{array}$

Substitute the $1\mathrm{m}$forx, and$3\times {10}^{6}m/s$in the above equation.

$$\begin{gathered} =\frac{1.0\mathrm{m}}{3.0\times {10}^6\mathrm{m}/\mathrm{s}} \\ =3.33\times {10}^{-7}\mathrm{s} \end{gathered}$$

Therefore, it will take $3.33\times {10}^{-7}\mathrm{s}$to travel$1.0\mathrm{m}$ for an electron in the horizontal direction.

Now, for the vertical motion of the electron, the initial vertical speed is given to be zero and the acceleration is gravitational acceleration directed downwards.

Hence, the second kinematical equation helps us determine the vertical distance traveled by the electron is the same time t.

$∆y=V_{i}t+\frac{1}{2}a{t}^2$

Substitute the 0 for ${\mathrm{V}}_{\mathrm{i}},9.8\mathrm{m}/{\mathrm{s}}^2\mathrm{for}\mathrm{a},\mathrm{and}3.33\times {10}^{-7}\mathrm{s}\mathrm{for}\mathrm{t}$in the above equation.

$$\begin{gathered} =0+\frac{1}{2}\times 9.8\mathrm{M}/{\mathrm{S}}^2\times {\left(3.33\times {10}^{-7}\mathrm{S}\right)}^2 \\ =5.4\times {10}^{-13}\mathrm{m} \end{gathered}$$

Therefore, the vertical distance traveled by the electron is $5.4\times {10}^{-13}\mathrm{m}$.

If the initial speed of the electron increases, it will take lesser time to travel the same horizontal distance of 1.0 m. Hence the time available for the vertical fall will also be less. It means that the vertical fall distance will also be reduced.