A ball is dropped from a top of a building. Describe the speed of the ball?

The speed is increasing. The acceleration of the ball is the gravitational acceleration (g).

9.8 metres per second (gavitational force)

Any object will accelerate from gravity -9.81 m/s^2. By integrating this acceleration over time, one obtains the velocity at any time t.

v = gt + v0. In this case, the ball started at rest, so the initial velocity v0 is zero.

By integrating velocity over time, one obtains the position y at any time t.

y = 1/2gt^2 + y0. We know that y0, the initial height of the ball, is 50 m.

So, when does the ball hit the ground (y = 0). Solve for t:

0 = 1/2 (-9.81 m/s^2) t^2 + 50 --> 100s^2 = 9.81 t^2 --> t = 3.2 s

You can answer the second question by examining the energy of the system. The kinetic energy when the ball hits the ground is determined by it s velocity. 1/2 mv^2. The velocity at t = 3.2s is v = g*3.2 = 10.2 m/s.

Uk = 52*m J/kg

The problem states that the ball looses half it s energy to the impact. So the new kinetic energy is 26*m J/kg. All of the kinetic energy will go to potential energy according to the equation Up = mgh. So,

Uk = Up --> 52m = mgh --> h = 52/g = 5.3 meters.