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 1.8 m/s . Part A How long after the release of the first stone does the second stone hit the water? Part B What was the initial speed of the second stone? Part C What is the speed of the first stone as it hits the water? Part D What is the speed of the second stone as it hits the water?

Answer #1

here,

the height of cliff , h0 = 56 m

t1 = 1 s

a)

the initial velocity of first stone , u1 = 1.8 m/s

let the time taken by first stone be t1

using first equation of motion

h0 = u1 * t1 + 0.5 * g * t1^2

56 = 1.8 * t1 + 0.5 * 9.81 * t1^2

solving for t1

t1 = 3.2 s

the time taken by second stone , t2 = t1 + 1 s = 3.2 s - 1 s = 2.2 s

b)

let the initial velocity of second stone be t2

using second equation of motion

h0 = u2 * t2 + 0.5 * g * t2^2

56 = u2 * 2.2 + 0.5 * 9.81 * 2.2^2

solving for u2

u2 = 14.7 m/s

the initial velocity of second stone is 14.7 m/s

c)

the speed of first stone when it hits the water , v1 = u1 + g * t1

v1 = 1.8 + 9.81 * 3.2 m/s

v1 = 33.2 m/s

d)

the speed of second stone when it hits the water , v2 = u2 + g * t2

v2 = 14 + 9.81 * 2.2 m/s

v2 = 35.58 m/s

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