Position, time physics problem!?

Question

A boat is moving at constant speed of 8m/s towards a bridge. You are standing on the bridge and wish to drop a package from rest onto the boat. Your hand is 15 meters above the boat deck. How far should the front of the boat be from the bridge so that the package will hit the deck?

(what i did was i found out the time it takes the package to fall 15 ft and how much the boat travels in that time. i just don't know how i'm supposed to find the distance of the boat from the bridge.)

thanks in advance!!!

Answer 1

h = vi + 1/2 gt^2

15 = 0 + 1/2 x 9.8 x t^2

t = sqrt(15x2/9.8) = 1.75 s

Distance should be just less than 1.75 x 8

Just less than 14 m, but more than 14 + length of deck

Answer 2

Your approach is correct. The basic requirement is that the boat needs to be exactly under the bridge when the packet reaches there. The packet has to travel 15 m. Assuming it merely falls at acceleration due to gravity (9.8 m/s^2),

Distance = (Initial Velocity*time) + 0.5*acceleration*time^2

15 = 0 + 0.5*9.8*time^2 = 4.9 * t^2

t^2 = 15/4.9 = 3.06

t = 1.75 s

so the dropped packet will take 1.75 s to reach where the boat would be.

The boat is moving at 8 m/s. In 1.75 s it will travel  8*1.75 = 14m

So the boat needs to be 14 m away from the bridge when you drop the packet. Did you get it? If not, drop me a note. good luck!