Aproton is projected into a uniform electric field that points vertically upward and has magnitude e. the initial velocity of the proton has a magnitude v0 and is directed at an angle a below the horizontal. (a) find the maximum distance hmax that the proton descends vertically below its initial elevation. you can ignore gravitational forces. (b) after what horizontal distance d does the proton return to its original elevation? (c) sketch the trajectory of the proton. (d) find the numerical values of hmax and dife = 500 n> c, v0 = 4.00 * 105 m> s, anda = 30.0°.

Part a)

here the proton is projected at some angle with the horizontal plate

so the two components of velocity is given as

now we will have acceleration due to electric force

now to find the maximum height we can use

here we will now find the maximum height from equation above

Part b)

total horizontal distance moved

now to find the time of whole motion we will have

now from above equation we have

Part c)

now if the given values are

now we have

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