In order to calculate the kinetic energy possessed by the 2 kg ball right before it hits the ground, we need to use the formula for kinetic energy, which is given by:
Kinetic Energy (KE) = 1/2 * mass * velocity^2
First, we need to determine the velocity of the ball right before it hits the ground. To do this, we can use the formula for gravitational potential energy:
Potential Energy (PE) = mass * gravity * height
Since the ball is dropped, initially it has no kinetic energy, only potential energy. Thus, the potential energy at the top of the building, when the ball is dropped, is equal to the kinetic energy right before it hits the ground. Therefore, we can set up the equation:
PE = KE
mass * gravity * height = 1/2 * mass * velocity^2
Here, the mass of the ball is given as 2 kg, the acceleration due to gravity is approximately 9.8 m/s^2, and the height from which the ball is dropped is 20 m. We can now solve for the velocity:
2 kg * 9.8 m/s^2 * 20 m = 1/2 * 2 kg * velocity^2
392 J = velocity^2
Taking the square root of both sides, we find:
velocity ≈ 19.8 m/s
Finally, we can substitute this value into the equation for kinetic energy:
KE = 1/2 * 2 kg * (19.8 m/s)^2
= 1/2 * 2 kg * 392.04 m^2/s^2
= 392.04 J
Therefore, the 2 kg ball will have approximately 392.04 Joules of kinetic energy right before it hits the ground.
essay 2- If a 2 kg ball is dropped from the top of a 20 m building, how much kinetic energy will it have right before it hits the ground?
1 answer