A mountain climber stands at the top of a 40.0 m cliff that overhangs a calm poo

Question

A mountain climber stands at the top of a 40.0 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 -1.00 m/s.

(a) How long after release of the first stone did the two stones hit the water? (Round your answer to at least two decimal places.) Incorrect: Your answer is incorrect. s

(b) What initial velocity must the second stone have had, given that they hit the water simultaneously? m/s

(c) What was the velocity of each stone at the instant it hit the water?

first stone:

second stone :

Super confused and I cant seem to get the right answer

Explanation / Answer

height of the cliff, h = 40 m
time interval between the two stones = 1 s = t
initial velocity of stone 1, u = 1 m/s downwards

time taken for stone 1 to reach the pool , t'
h = ut + 0.5*gt^2
40 = t' + 0.5g*t'^2
4.9t'^2 + t' - 40 = 0
solving for t'
t' = 2.756 s

a. so the two stoens hit water after 5.922 s
b. let the second stone have initial velocity u
   40 = u(2.756-1) + 0.5g(2.756-1)^2
   u = 14.1746 m/s
   so the second stone should have a velocity of 14.1746 m/s in downward direction for the two stones to splash together
c. stone 1, v = 1 + 9.81*2.756 = m/s
   stone 2, v = 14.1746 + 9.81*1.756 = 31.40096 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.