I have no idea how to solve this please help Protons are pro

I have no idea how to solve this please help

Protons are projected with an initial speed v_i = 9.55 x 10^3 m/s into a region where a uniform electric field E = (-720j) N/C is present, as shown in the figure above. The protons are to hit a target that lies at a horizontal distance of 1.27 mm from the point where the protons are launched. Find (a) the two projection angles Theta that result in a hit and (b) the total lime of flight for each trajectory. 36.9 Degree, 53.1 Degree, 166 ns, 221 ns

Solution

The motion of the protons can be considered as similar to that of a projectile. Here, the protons will suffer a constant downward force of qE due to the electric field of 720 N/C acting in the given area. Hence will have an acceleration towards the negative Y axis of qE/m

The magnitude of that acceleration would be:

a = 1.60217662 × 10^-19 x 720 / 1.6726219 × 10-27 =  689.676x 10^8 m/s^2

However, they will not suffer any acceleration along the horizontal direction and will continue to travel with the same horizontal velocity, VCos throughout the motion.

We can write the equation of motion along the vertcial direction as:

- V Sin = V Sin - aT

or T = 2VSin / a

For the motion along the horizontal direction, as the speed remains constant horizontally,

1.27 x 10^-3 = 2Sin Cos V^2 / a = V^2Sin2 /a [Since 2SinCos = Sin2]

or 1.27 x 10^-3 = 9.55 x 9.55 x 10^6 Sin2 / a

Sin2 = 689.676x 10^8 x 1.27 x 10^-3 / 9.55 x 9.55 x 10^6 = 0.9604

Hence 2 can be around 74 degrees or 106 degrees

Therefore can be 37 degrees of 53 degrees.

b.) Now,Time of flight as mentioned in the part above would be:

T = 2VSin / a = 2 x 9.55 x 10^3 x 0.60042 / 689.676 x 10^8 = 0.01663 x 10^-5 = 166.3 ns

Also for = 53.1, T = 2 x 9.55 x 10^3 x 0.7997 / 689.676 x 10^8 = 221 ns