
#1
Feb2410, 11:22 PM

P: 18

1. The problem statement, all variables and given/known data
A mountain climber stands at the top of a 50.0m cliff hanging over a calm pool of water. The climber throws two stones vertically 1.0 s apart and observes that they cause a single splash when they hit the water. The first stone has an initial velocity of + 2.0 m/s. a) How long after release of the first stone will the two stones hit the water? b) What is the initial velocity of the second stone when it is thrown? c) What will the velocity of each stone be at the instant both stones hit the water? 2. Relevant equations displacement = V_{1}time + 1/2atime^{2} V_{2} = V_{1} + a(time) 3. The attempt at a solution For a, I'm essentially, interpreting it as the time is takes the first stone to hit the water, as that is the time after the release. And the second stone is thrown during this time. 50.0m for displacement, +2.0 for velocity 1, acceleration is 9.81. I rearrange and use the quad formula for solve to get the time as 3.4s. Is my interpretation of part a correct? For b, I subtract 1.0s from the 3.4s, as the second is released 1.0s after. Same equation used in part a, to solve for V1, which turns out as 9.06 m/s. Number seems somewhat reasonable, since it falls much faster. For c, I solve the final velocities seperately, both using the formula for V2, but for differnet times and init. velocities. Are my thought processes and method correct? 



#2
Feb2510, 07:12 AM

P: 32

Hi!
Yes, you are going right. 


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