a 10kg ball is atop a 20m hill. what is the speed when it reaches the bottom how did you get the anwser
To find the speed of the ball when it reaches the bottom of the hill, you can use the conservation of mechanical energy. The potential energy of the ball at the top of the hill is converted into kinetic energy at the bottom.
The potential energy (PE) of an object is given by the equation PE = mgh, where m is the mass (10 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the hill (20 m).
So, the potential energy at the top of the hill is PE = 10 kg * 9.8 m/s^2 * 20 m = 1960 J.
According to the conservation of mechanical energy, this potential energy is converted entirely into kinetic energy (KE) at the bottom of the hill. The kinetic energy equation is KE = (1/2)mv^2, where m is the mass of the ball (10 kg) and v is the velocity.
Equating the potential energy and kinetic energy, we have:
KE = PE
(1/2)mv^2 = 1960 J
Now we can solve for v:
v^2 = (2 * 1960 J) / 10 kg
v^2 = 392 m^2/s^2
v ≈ 19.8 m/s
Therefore, the speed of the ball when it reaches the bottom of the hill is approximately 19.8 m/s.
Keep in mind that this calculation assumes no energy losses due to air resistance or friction. In reality, the actual speed might be slightly lower due to these factors.