A rock climber stands on top of a 56 m -high cliff overhanging a pool of water.

Question

A rock climber stands on top of a 56 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s apart and observes that they cause a single splash. The initial speed of the first stone was 2.4 m/s . please show answers and steps to answer! THANKS!

A.  How long after the release of the first stone does the second stone hit the water?                                  

B. What is the speed of the second stone as it hits the water?

C. What was the initial speed of the second stone?                                                                                                                           

D. What is the speed of the first stone as it hits the water?

Explanation / Answer

here,

a)

time taken by first stone be t1

h = u1 * t1 + 0.5 * g * t1^2

56 = 2.4 * t1 + 0.5 * 9.81 * t1^2

solving for t1

t1 = 3.14 s

the time taken by the seccond stone is (t1 - 1 = 2.14 s)

b)

let the initial speed of seccond stone be u2

h = u2 * t2 + 0.5 * g * t2^2

56 = u2 * 2.14 + 0.5 * 9.81 * 2.14^2

u2 = 15.7 m/s

the final speed of seccond stone , v2 = u2 + g * t2

v2 = 15.7 + 9.81 * 2.14 = 36.7 m/s

c)

the initial speed of stone2 is 15.7 m/s

d)

the final speed of first stone , v1 = u1 + g * t1

v1 = 2.4 + 9.8 * 3.14 = 33.2 m/s

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