# Physics homework help

MC363A
i need help with this problem. I've tried for a while now and it is driving me crazy! I've tried about everything including using what little calculus i remember from last year! please help! the problem is: you are standing on a cliff and you drop a rock straight down, neglecting air resistance. you hear the rock hit the water 3.4 seconds later, if the speed of sound is 340 m/s, how high is the cliff. i would greatly appreciate any help, and as i have hit 2 or 3 dead ends that filled sheets of paper, i see it as pointless to put it on here. just trust me when i say i need a little insight. i tried to set it up as such: t = delta X / V = sqrt(2 deltaX/g) is there any truth to this equation?

## Answers and Replies

stunner5000pt
i'm not quite sure how you got that equation

anyway here's how i would solve it
for the speed of sound to travel back up the distance traveled was teh height of the cliff so $$d = 340 t_{sound}$$
now for the fall
v1 = 0
v2 = ?
a = -9.8m/s^2 (gravity points down, i took down to be negative, it doesn't matter whichever way you take it )
$t_{fall} = ?$
now relate what u have to what you don't have. Also what u don't need can be omitted in that equation you form. How are $$t_{fall} and t_{sound}$$ related? Tfall is the time it took for hte stone to fall into the water. Tsound is the time it took for the sound to reach you. How long was this entire trip?

MC363A
thats where i got my equation from. d = 1/2at1^2 therefore, t1 = sqrt(2d/g) and t2 = d/v(sound) t1+t2=t(total) therefore, (d/v)+sqrt(2d/g) = t(total) it's the rest of it i get messed up on, I've tried it severl times, and it appears you are hinting at this equation, a little insight?

stunner5000pt
this is how i did it
$$d = v_{1} t + \frac{1}{2} at_{fall}^2$$
v1 = 0 so
$$d = \frac{1}{2} (-9.8) t_{fall}^2$$...1
the speed of sound part
$$d = v_{sound} t_{sound}$$
$$d = 340 t_{sound}$$...2
so
$$t_{sound} + t_{fall} = 3.4$$...3

combine 1,2, and 3 and should should be able to solve for one of the times. Thus u can solve for the distance.