please answer the follwing questions 1 a particles of mass 2

please answer the follwing questions

1- a particles of mass .20 kg and charge 25.0micro c has an intial velocity of 20.0 m/s in the x- direction and 5.0 m/s in the postive y-direction . the particles enter a uniform electric filed pointed downward in the negtive y- direction . if the particles\'s intial height above the ground is15 meter .how far will the particles travel before it strikes the ground ? E=2.0 *10^4 N/C . Do not ignore gravity.

2-it is the same question above exept cherge is 2.5

Solution

the mass of the charge is M=0.2 kg

the charge it holds q=25 uC= 25X10^-6 C

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

direction along yaxis Vy_{= 5.0 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-25 X 10^-6=-0.5N

acceleartion due to field = F_e/M= -0.5/0.2= -2.5 m/s²

net acceleration = a_e+g = -10 +(-2.5) = -12.5

^{for distance travelled 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 =-12.5m/s² (as calculated) and u =5 m/s

-15=5t+(-12.5)t²/2

solving the above quadratic we get t=2 s

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= 20X2 =40m

2.

the mass of the charge is M=0.2 kg

the charge it holds q=2.5 uC= 2.5X10^-6 C

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

direction along yaxis Vy_{= 5.0 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-2.5 X 10^-6=-0.05N

acceleartion due to field = F_e/M= -0.05/0.2= -.25 m/s²

net acceleration = a_e+g = -10 +(-.25) = -10.25

^{for distance travelled 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 =-10.25m/s² (as calculated) and u =5 m/s

-15=5t+(-10.25)t²/2

solving the above quadratic we get t=1.29 s

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= 20X1.29 =25.82m