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Ann [662]
3 years ago
12

A rock climber stands on top of a 50 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s a

part and observes that they cause a single splash. The initial speed of the first stone was 2.4 m/s .
a) How long after the release of the first stone does the second stone hit the water?
b) What was the initial speed of the second stone?
c) What is the speed of the first stone as it hits the water?
d) What is the speed of the second stone as it hits the water?
Physics
1 answer:
bagirrra123 [75]3 years ago
3 0
<h2><em>Answer: b) What was the initial speed of the second stone?</em></h2>

Explanation:

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A billiard ball is dropped from a height of 64 feet. Use the position function s(t) = –16???? 2 + ????0???? + ????0 to answer th
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Answer:

s(t) = -16*t^2 + 64

v(t) = -32*t

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v(t) = 64 ft/s ... At impact

Explanation:

Given:-

- The height of the billiard ball t = 0 , h = 64 ft.

- The position function of an object under gravity is given by:

                                    s(t) = -16*t^2 + v_o*t + s_o

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a. Determine the position function s(t),

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- To determine the position function we must initialize our problem and use the given general equation.

- s(t) is the position of the billiard ball from the ground at time t. So when t = 0, then s(t) = h. Hence, we have:

                                  s(t) = s_o = h = 64 ft

- Similarly we know that v_o is the initial velocity of the ball. Since, the ball was dropped we say that the initial velocity v_o = 0. Hence, the position of the ball from ground is given by following expression:

                                  s(t) = -16*t^2 + 64  

- To find the velocity expression v(t) we will take the time derivative of the position expression s(t) as follows:

                                  v(t) = d s(t) / dt

                                  v(t) = -16*2*t + 0

                                  v(t) = -32*t ft/s

- Similarly, the expression for acceleration a(t) is given by the time derivative of the velocity expression v(t) as follows:

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                                  a(t) = -32*t

                                  a(t) = -32 ft/s^2

- The velocity of ball at impact can be determined by evaluating s(t) = 0 and find the value for time t. Then that time t can be substituted in the velocity expression v(t) for final velocity. Or we could use the following 3rd kinematic equation as follows:

                                 v(t)^2 - 0^2 = 2*a(t)*s_o

                                 v(t)^2 = 2*(32)*(64)

                                 v(t) = 64 ft/s

- The ball has a velocity of 64 ft/s at impact!

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