Projectile Motion involving eletric fields

[SF2K4]

Homework Statement

Ok, so, I've about had it rattling my brain on this. I'm stumped.

Q:

Protons are projected with an initial speed vi = 9.89x10^3 m/s into a region where a uniform electric field E = (-720 j) N/C is present. The protons are to hit a target that lies at a horizontal distance of 1.27 mm from the point where the protons cross the plane and enter the electric field.

Find the two projection angles that result in a hit and the corresponding times of flight.

Homework Equations

The only thing I was able to compute (hopefully I got it right) was that the acceleration in the y-direction (as there is none in the x) is -6.89687x10^10 m/s^s. Since E is -720j and knowing the charge and mass of a proton I calculated that... but that's about all I've got.

I know all of my kinematic equations but I'm still at a loss on this problem...

The Attempt at a Solution

...

 

[mukundpa]

Think
What is the the net vertical distance traveled by the proton before hitting the target?

Will this help to find time elepsed?

What is the horizontal distance traveled during this time?

 

[SF2K4]

The horizontal distance is 1.27mm... but there are two vertical distances depending on the angle theta...

 

[chanvincent]

decompose the problem into x and y direction... write down the equation of motion in terms
of the angle theta and time t...

For the x direction, the distance travel is 1.27mm... v_x depend on the angle theta and the
projection velocity... it is easy, you should have no problem on this one...
For the y direction, vertical distance travel is zero... right?

Now you have two equation with two unknown...

 

[SF2K4]

The equations I have are:

dx = vxit
vyf = vyi + at
dy = vyit + (1/2)at^2

However, t is unknown and vyi and vxi are all dependent on theta...

 

[chanvincent]

so, you have
$$d_x = v_x^i t$$
$$d_y = v_y^i t + \frac{1}{2} at^2$$

and from the relationship

$$v_x^i = v^i cos\theta$$
$$v_y^i = v^i sin\theta$$

The only unknown(s) are t and $$\theta$$
Could you solve for $$\theta$$ by eliminating t??

 

[SF2K4]

But then wouldn't I be left with something like:

$$d_x = v^i cos\theta t$$

Which would result in:

$$t = \frac{d_x}{v^i cos\theta}$$

And after using $$sin^2\theta + cos^2\theta = 1$$ I should be able to use a quadratic formula to get two values of t... but I only end up getting one and it doesn't result in a correct answer.

 

[chanvincent]

Why would you use $$sin^2\theta + cos^2\theta = 1$$ ?

since you know $$t = \frac{d_x}{v^i cos\theta}$$, just plug it into
$$d_y = v^i sin\theta t + \frac{1}{2} at^2$$

with the t eliminated, now you are left with one equation with one unknown...
Everything shall be crystal clear from here, right?