Projectile Motion Word Problem

[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.

 

[nlightner]

Yes, you are on the right track for both attempts. However, I would suggest using the second attempt as it takes into account the acceleration due to gravity, which is necessary for accurate calculations in projectile motion problems. Your calculations show that a horizontal speed of 3.61 m/s would be needed to reach tower 2 in 5.53 seconds, which is well below the maximum velocity of 5 m/s stated by the defense attorney. This supports the prosecution's argument that the defendant did not have to throw the money bag with a velocity greater than 5 m/s to reach tower 2. Additionally, you could also use the Pythagorean theorem to calculate the actual distance the money bag traveled in the air, which would be slightly longer than the horizontal distance of 20 m. This further supports the prosecution's argument that the defendant could have thrown the money bag with a lower velocity and it would still reach tower 2. Overall, your calculations and reasoning provide strong evidence for the prosecution's case.