Can the Defendant Throw the Money Bag Under 5 m/s?

[georgiaa]

Homework Statement

From the roof of tower 1, a theif throws a money bag to an accomplice on the roof of tower 2, which is just west of tower 1. The two towers are separated by a mall. The defense attorney contends that in order to reach the roof of tower 2, the defendant would have had to throw the money bag with a maximum velocity of no more than 5m/s. Tower 1 is 250 m high, tower 2 is 100m high and the mall is 20m wide. How will you advise the prosecuting attorney?

Homework Equations

The big 5 equations
v= d/t
Cosine Law
Sine Law
Pythorean

The Attempt at a Solution

First attempt:
I used calculations assuming a 45 degree angle to the horizontal, then determined time using a big 5 equation. I then used this time to determine that the money traveled a total distance of 31.5 m.

Second attempt
d = 1/2 a t^2
-150 = .5(-9.81) t^2

5.53 sec is the time to fall from tower 1 to 2

d = vt
20 = v(5.53)
v = 3.61 sec is horizontal speed needed to get to tower 2 in 5.53 sec

Am I on the right track for either attempt?

 

[LowlyPion]

Pretty much. If you have 5.5 sec in the air before it reaches 150 m, then to go 20 m in that time is plenty to spare if they can throw at 5 m/s.

Try beaten by parents as a child for a defense instead.