So I'm doing a simple Physics homework problem, and I'm honestly unsure if I'm using the proper formulas for each step. The problem is as follows:

A key falls from a bridge that is 47 m above the water. It falls directly into a model boat, moving with constant velocity , that is 15 m from the point of impact when the key is released. What is the speed of the boat?

I used the intial formula (to find the time it takes for the key to fall) of: $x-x_{0} = v_{0}t+\frac{1}{2}at^{2}$ and the result I got was about 3.10 seconds (it was actually 3.097...but I rounded)

Then I used the formula $x-x_{0} = \frac{1}{2}(v_{0}+v)t$ to solve for velocity of the boat - getting 9.7 m/s, but apparently this answer is incorrect. I'm sure it's probably because I'm not using the proper formula for the second step (of finding the speed of the boat) but I'm not sure which one to use (as the only information I have is time and distance). Could anyone point me in the right direction?

  • $\begingroup$ The boat is 15 meters horizontally away from the point where the key is going to reach either the water or the deck of the boat. The boat has 3.1 seconds to get there to intercept the key, so it doesn't end up in the water. How fast would the boat need to go to make the interception? $\endgroup$ – colormegone Jan 29 '15 at 3:55
  • $\begingroup$ As a note about the formula you show, the distance the boat needs to cover would be $ \ x \ - \ x_0 \ = 15 \ $ meters, and since the speed of the boat is taken to be constant, $ \ v_0 \ = \ v \ $ , so the right-hand side of the equation reduces to $ \ \frac{1}{2} (v_0 + v_0) t \ = \ v_0 \ t \ $ . $\endgroup$ – colormegone Jan 29 '15 at 4:02

You found the time correctly. Now, in that same time, the boat travels $15$m. So, $$x=vt$$ $$15=v(3.1)$$

You get $v=4.84$m/s. Is that correct?

  • $\begingroup$ I honestly had no idea it was that simple, I was staring at the constant acceleration equations trying to figure out what I was doing wrong. Thank you so much! $\endgroup$ – secondubly Jan 29 '15 at 3:59
  • $\begingroup$ You're welcome. :-) $\endgroup$ – Tejas Jan 29 '15 at 4:02

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