A charged particle of mass m of 1 kg and charge q = 2 uC is thrown from a horizontal ground at an angle teta = 45 degree with speed 20 m/s. In space a horizontal electric field E = 2 x 10^7 V/m exist. Find the range on horizontal ground of the projectile thrown

Answer 1

v(ox) =v(o)•cosθ =20•0.707=14.14 m/s,
v(oy) =v(o)•sinθ =20•0.707=14.14 m/s.
The motion from the initial point to the highest point is accelerated in x-direction and decelerated in y-direction:
x: x=v(ox)•t+a•t^2/2
v(x) = v(ox) +a•t
y: y=v(oy)•t - g•t^2/2
v(y) = v(oy) - g•t.
For the highest point
v(y) = 0, t = v(oy)/g = v(o)•sinθ/g = 20•0.707/9.8 = 1.443 s.
ma = qE a = qE/m = 2•10^-6•2•10^7/1 = 40 m/s^2
L1 = v(ox)•t+a•t^2/2 =14.14•1.443 +40•(1.443)^2/2 =62.04 m.
v(x) = v(ox) +a•t =14.14 + 20•1.443 = 43 m/s.
This speed is the initial speed for the second part of horizontal motion
V(o) = 43 m/s.
The motions in vertical and horizontal directions are independent from each other, therefore, for vertical motion the time of descent is t =1.443 s.
For the motion from the highest point to the final point
L2= V(o)•t+a•t^2/2 =43•1.443 + 40•(1.443)^2/2 =103.7 m.
L = L1 +L2 = 62.04 + 103.7 = 165.74 m.