Not really sure how do this problem or even where tobegin... An electrically cha

Question

Not really sure how do this problem or even where tobegin...
An electrically charged point particle that has a mass of 1.5x 10^-6 kg is launched into a uniform electric field with amagnitude of 925 N/C. The particle'e initial velocity is 8.80m/s,directed eastward, and the external electric field is also directedeastward. A short time after ebing launched, the particle is 6.71 x10^-3 m below and 0.160m east of its initial position. Determinethe net charge carried by the particle, including the algebraicsign (+ or -). Not really sure how do this problem or even where tobegin...
An electrically charged point particle that has a mass of 1.5x 10^-6 kg is launched into a uniform electric field with amagnitude of 925 N/C. The particle'e initial velocity is 8.80m/s,directed eastward, and the external electric field is also directedeastward. A short time after ebing launched, the particle is 6.71 x10^-3 m below and 0.160m east of its initial position. Determinethe net charge carried by the particle, including the algebraicsign (+ or -).

Explanation / Answer

An electrically charged point particle that has a mass of 1.5x 10^-6 kg is launched into a uniform electric field with amagnitude of 925 N/C. The particle's initial velocity is 8.80m/s,directed eastward, and the external electric field is also directedeastward. A short time after being launched, the particle is 6.71 x10^-3 m below and 0.160m east of its initial position. Determinethe net charge carried by the particle, including the algebraicsign (+ or -).
the time taken by the particle to reach the point below=t
using the equation
h=0+1/2g*t*t
6.71*10-3=0+4.9*t*t
t=3.70*10-2sec
in the same time the particle reaches a horizontal distance of0.160m
let the electric charge present on the particle be q
the force due to electric field=Eq=F=925q
so 0.160=8.8*3.70*10-2+1/2 (F/mass)(3.70*10-2)2=
      0.160  =8.8*3.70*10-2+1/2*925q /1.5 x10-6   *(3.70*10-2)2
calculating we will get the magnitude of the charge.
as the charge is moved in the direction of the field the charge ispositive.

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