A ball is dropped from the top of a tall building of height = 50 m. About how long does it take for the ball to hit the ground? (Neglect air resistance) b) If the ball has lost half the magnitude of its impact momentum immediately after it re-coils, to what height does the ball reach afterits first rebound?

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Any object will accellerate due to 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...

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Any object will accellerate due to 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. How high will it go? 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.

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