A tennis ball is dropped from 1.24 m above
the ground. It rebounds to a height of 1.05 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s
2
. (Let
down be negative.)
Answer in units of m/s
21
To find the velocity at which the tennis ball hits the ground, we can use the principle of conservation of energy. The total mechanical energy of the ball at the top (when it is dropped) is equal to the mechanical energy at the bottom (when it hits the ground).
The mechanical energy of an object includes both its potential energy and its kinetic energy. At the top, the ball has potential energy due to its height above the ground, and no kinetic energy. At the bottom, all potential energy is converted into kinetic energy.
The potential energy of an object is given by the equation PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height. Kinetic energy is given by the equation KE = 0.5mv^2, where v is the velocity.
Since the height at the top is 1.24 m and at the bottom is 1.05 m, the change in height is 1.24 - 1.05 = 0.19 m. The mass of the tennis ball is not given, so we can assume it to be negligible as compared to h and g. Thus, we can neglect the mass in our calculations.
At the top, the potential energy is PE = mgh = 0.19 * 9.8 = 1.862 J.
At the bottom, the potential energy is zero because the height is zero. Therefore, at the bottom, all the potential energy is converted into kinetic energy.
So, 1.862 J of potential energy is equal to the kinetic energy.
Therefore, 1.862 J = 0.5mv^2.
We can now solve this equation to find the velocity, v.
1.862 = 0.5 * v^2.
Rearranging the equation, we get v^2 = 1.862 / 0.5.
Taking the square root of both sides gives v ≈ √(1.862 / 0.5).
Calculating the value, we find v ≈ 2.72 m/s.
Thus, the velocity at which the tennis ball hits the ground is approximately 2.72 m/s.