Question

On planet Tehar, the free-fall acceleration is the same as that on Earth, but there is also a strong downward electric field that is uniform close to the planet's surface. A1.80-kg ball having a charge of 5.22 mC is thrown upward at a speed of 20.1 m/s. It hits the ground after an interval of 2.10 s. What is the potential difference between the starting point and the top point of the trajectory? (Use 9.80 m/s² for the acceleration due to gravity.) (In kV) (sig figs count)

Answer 1

According to the concept of electric potential

K.E=qV

1/2m(v^2-u^2)=qV

Now we find the final velocity of ball

V=u+gt

V=20.1+9.8*2.1=40.2 m/s

Now we find the electric potential difference

Electric potential difference V

=(1/2*1.8*(40.2^2-20.1^2))/5.22*10^-3

=208.97*10^3 v