A mountain climber stands at the top of a 45.0 m cliff that overhangs a calm pool of water. She throws two stones vertically downward 1.00 s apart and observes that they cause a single splash.The first stone had an initial velocity of -1.40 m/s.
(a) How long after release of the first stone did the two stones hit the water?
(b) What initial velocity must the second stone have had, given that they hit the water simultaneously?
(c) What was the velocity of each stone at the instant it hit the water?
I got the answer for part a correct so it should be 2.89 seconds.
I used the formula delta x = V0t + 1/2at^2 for part b to find the initial velocity for the 2nd stone.
45 m = V0 (2.89) + (1/2) (-9.8) (2.89)^2
45 m = V0 (2.89) + (-4.9) (8.3521)
45 m = V0 (2.89) - 40.93
85.93 = V0 (2.89)
This is what I got as answer for part b but I doubt it is right because that is too large of a number. Can some1 plz help? ty
never mind I see what I did wrong.
no I dont; its still wrong :P
o I get it now.....thanks!
- Anonymous1 decade agoFavorite Answer
That 45m needs to be NEGATIVE 45m if your acceleration is -9.8 (ie you are calling down negative).
Of course this means that V0 will be negative for the second rock (ie thrown downward) just as it was for the first.
Also, note that the time should be 1.89, not 2.89, because the second stone is released later.