   Chapter 2, Problem 68AP

Chapter
Section
Textbook Problem

A mountain climber stands at the top of a 50.0-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 −2.00 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?

(a)

To determine
The time for this stone to hit the water.

Explanation

Given Info:

The initial velocity of the first stone is 2.00m/s

The acceleration of the stone is 9.80m/s2

The displacement of the first stone is 5.00m

The formula used to calculate the velocity at which the first stone hit water is,

v1=v012+2aΔy

• Δy is the displacement of the first stone
• v1 is the final velocity of the stone
• v01 is the initial velocity of the stone
• a is the acceleration

Substitute 2.00m/s for v01 , 9.80m/s2 for a and 5.00m for Δy to find v1 .

v1=(2.00m/s)2+2(9.80m/s2)(5.00m)=31.4m/s

Thus, the velocity at which the stone hits the water is 31

(b)

To determine
The initial velocity of the second stone.

(c)

To determine
The final velocity of each stone.

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