Aball is thrown upward from the ground. at a helght of 10 meters above the ground, the ball has a potential energy of 50 joules (with the potential eneray equal to zero at ground level) and is moving upward with a kinetic energy of 50 joules. air friction is negligible. the maximum height above ground reached by the ball is most nearly

a. 10 m
b. 20m
c. 30 m
d. 40 m
e. 50 m

h = 20 m

Explanation:

height (h) = 10 m

potential energy (PE) = 50 J

kinetic energy (KE) = 50 J

what is the maximum height reached by the ball

At 10 meters above the ground  potential energy (PE) = 50 J

         mgh = 50

         recall that height (h) = 10

          mg (10) = 50

          mg  = 5 .....equation 1

The total energy of the system at 10 meters is given as PE + KE = 100 Joules. From the conservation of energy the total energy in the system still remains the same at the maximum height, this is because energy can neither be created nor destroyed but can be transformed from one form to another. At the maximum height the velocity (v) of the system is 0, which means all its kinetic energy (KE) is converted to potential energy. Therefore the system consist only of potential energy.

          KE + PE = 100

          + mgh = 100 J

         velocity (v) = 0 at maximum height therefore

         0 + mgh = 100

         mgh = 100

         recall that mg = 5 from equation 1, therefore

         5h = 100

         h = 20 m

The answer is 4f======since electrostatic force is given by    we let f be the electrostatic force when  r = 1 m    if we let  denote the electrostatic force when the distance is decreased so that r = 0.5m, then    we established earlier that f = kq₁q₂. therefore    the force will now be 4f.