Aboy releases a small 1 kg toy truck from rest down a 5 meter high hill. assuming there is no friction, what is the maximum kinetic energy the truck could have at the bottom?

Potential energy is the energy possessed by the body while at rest while the kinetic energy is the energy possessed by a body in motion. Potential energy is given by Mgh, where m is the mass, g is the gravitational acceleration and h is the height.
kinetic energy is given by 1/2 mv², where m is the mass and v is the velocity of the body. According to the law of conservation of energy , energy can not be destroyed nor created but cannot be created. Therefore in this case the potential of the body at rest was transformed to kinetic energy and thus the kinetic energy will be equal to the potential energy.
Hence, energy = 1 ×5×10 = 50 Joules
 The kinetic energy will be 50 Joules

First, determine the speed of the truck after travelling 5 meters. This can be calculated through the equation,

     2ad = Vf² - Vi²

a is equal to the acceleration due to gravity of 9.8 m/s², d is the distance equal to 5 m, and Vi is zero for free-falling object. Substituting the known values,

   2(9.8)(5) = Vf²

The value of Vf from the equation is equal to 9.90 m/s. 

The equation for the kinetic energy is,
          KE = 0.5mv²

Substituting the known values,
          KE = 0.5(1 kg)(9.90 m/s)²
           KE = 49 J

49 J

Iagree c would be the answer