# Is there a simpler way to solve this free fall problem?

1. Sep 12, 2007

### novice@physics

1. The problem statement, all variables and given/known data
(see scanned picture)

2. Relevant equations
kinematic equations of motion (*under condition that acceleration is constant)

3. The attempt at a solution
(see scanned picture)

i was just wondering if there is a more simpler way to attack this problem. only after an hour later (and constantly looking at the equations i can work with) did i have some weird flash of insight to do what i did and just jot things down on the paper until i arrived at the correct answer.

the thing with free fall problems (or any physics problems for that matter) is that it is usually useless to "memorize" a method on solving a particular problem because not all the problems of a certain topic will be the same (i.e. different wording, certain conditions, etc.). that means, if i saw this problem a week later (assume that i did not think about this problem ever since) i would probably be confused and spend an hour again until i get that flash of insight, if at all. not sure if this is normal for you guys, but seems inefficient, especially when you have more homework problems to solve. i can only already see myself on a test day being like this.

just to let you know, the problems in my book have a difficulty range of 1 dot to 3 dots. this was a 2 dot problem. (sad face)

2. Sep 13, 2007

### learningphysics

Hint: The displacement in the last second... is the displacement over t seconds (assuming t is the time it takes to fall) - displacement over the first (t-1) seconds.

That gives you a way to find the time t of the fall... then you can get h.

3. Sep 14, 2007

### dynamicsolo

Because of the sort of information given in the problem, there isn't a good way to avoid needing to solve for the time; the "velocity-squared" equation won't tell you enough. learningphysics's hint is essentially what you need to do. But you can simplify the work by saying that after t seconds, (1/2)g(t^2) = (h-38) and that after one more second, (1/2)g([t+1]^2) = h . If you substitute the second equation into the first and multiply out the binomial-square, the (1/2)g(t^2) cancels out and you only need to solve a *linear* equation for t . Putting this value into your second equation then gives h.