Answer
$\theta_0=16^o$
Work Step by Step
This problem is similar to a question concerning a particle moving through the air because the force, and therefore acceleration, is constant. To calculate angle using the kinematics equations, we need to determine initial velocity.
$F=qE=-eE=-6.09\times10^{-16}N$
Using $F=ma$, $a=\frac{-6.09\times10^{-16}N}{9.11\times10^{-31}kg}=6.68\times10^{14}m/s^2$
The electron has constant horizontal velocity, so $v=\frac{7.2cm}{t}$ and similarly, $t=\frac{7.2cm}{v_i\cos(\theta_0)}$.
Using $v_f=v_i+at$, at the top of its motion, $0m/s=v_{i}\sin(\theta_0)+(6.68\times10^{14}m/s^2)(\frac{7.2cm}{v_i\cos(\theta_0)})$
Using $y_f=y_i+v_{iy}t+\frac{1}{2}at^2$,
$0.50cm=y_i+v_i\sin(\theta_0)(\frac{7.2cm}{v_i\cos(\theta_0)})+\frac{1}{2}(6.68\times10^{14}m/s^2)(\frac{7.2cm}{v_i\cos(\theta_0)})^2$
Solving the system of two equations, we get $\theta_0=\arctan\big(\frac{2h}{d}\big)$
$\theta_0=16^o$
