The Police Department has hired you as a consultant in a robbery investigation. A thief allegedly robbed a bank and, to escape t
he pursing security guards, took the express elevator to the roof of the building. Then, in order to not be caught with the evidence, the thief allegedly threw the money bag to a waiting accomplice on the roof of the next building. The defense attorney contends that in order to reach the roof of that next building, the defendant would have had to throw the money bag horizontally with a minimum velocity of 10 meters/second. However, in a test, the accused could throw the bag with a maximum horizontal velocity of no more than 5 meters/second. How will you advise the prosecuting attorney
The vertical distance between the roofs of the two buildings is 150 meters. Let’s use the equation below to determine the time required for the bag to drop 150 meters.
d = vi * t + ½ * 9.8 * t^2, d = 150 m Since the bag is thrown horizontally, the initial vertical velocity is 0 m/s. 150 = ½ * 9.8 * t^2, t = √(150/4.9) ≈ 5.53 seconds
We can use the time to determine the minimum horizontal velocity required for the bag to travel 20 meters. v = 20 ÷ √(150/4.9) ≈ 3.61 m/s
Since this velocity is less than 5 m/s. I believe that the bag to accused could throw the bag to the accomplice. I would advise the prosecuting attorney to do some pleas bargaining to get the accused the best deal possible. The accused is GUILTY!
