A particle of mass 020 kg and charge 250 uc has an initial v

A particle of mass 0.20 kg and charge 25.0 uc has an initial velocity of 20.0 m/s in the x direction and 5.0 m/s in the positive y-direction. The partical enters a uniform electric files points downward in negative y-direction. If the partical\'s initial height above ground is 15.0 meters, how far will the particle travel before it strikes the ground? E = 2.0 times 10^4 N/C note do not ignore gravity.

Solution

the mass of the charge is M=0.2 kg

the charge it holds q=250 uC= 250X10^-6 C

direction along x axis V_{x= 20 m/s}

direction along yaxis Vy_{= 50 m/s}

electric field along negative y direction = 2X10⁴C

acceleration due to gravity = 10m/s²

force due to electric field F_e=qXE=2X10⁴ X-250 X 10^-6=-5N

acceleartion due to field = F_e/M= -5/0.2= -25 m/s²

net acceleration = a_e+g = -10 +(-25) = -35m/s²

^{for distance traveelled fr an accelearitng particle the formula we use will be}

y=ut+at²/2

where y is the distance , u initial velocity , a is the acceleration and t the time taken

here y =15m and a =-35m/s² (as calculated) and u =50 m/s

-15=50t+(-35)t²/2

solving the above quadratic we get t=0.27s

that is the time taken by the charge to hit the ground

so the distance travelled along x direction would be

d= V_x X t= 20X0.27 =5.4 m