A body of mass 5 kg is thrown vertically up with a kinetic energy of 490 J. The height at which the kinetic energy of the body becomes half of the original value is

To solve this problem, we need to use the principles of energy conservation and the relationship between kinetic energy and potential energy.

Step 1: Calculate the initial velocity of the body We know that the kinetic energy (KE) of a body is given by the formula KE = 1/2 * m * v^2, where m is the mass and v is the velocity. We can rearrange this formula to solve for v: v = sqrt((2KE)/m). Substituting the given values, we get v = sqrt((2490 J)/5 kg) = 14.97 m/s.

Step 2: Calculate the height at which the kinetic energy becomes half When the kinetic energy of the body becomes half of the original value, the remaining energy is converted into potential energy (PE). The potential energy is given by the formula PE = m * g * h, where m is the mass, g is the acceleration due to gravity (approximated as 9.8 m/s^2), and h is the height. We can rearrange this formula to solve for h: h = PE/(m*g).

The potential energy at this point is equal to the original kinetic energy minus the current kinetic energy, which is half the original kinetic energy. So, PE = 490 J - 1/2 * 490 J = 245 J.

Substituting the given values into the formula for h, we get h = 245 J / (5 kg * 9.8 m/s^2) = 5 m.

So, the height at which the kinetic energy of the body becomes half of the original value is 5 meters.