A rock climber stands on top of a 59 m -high cliff overhanging a pool of water.

Question

A rock climber stands on top of a 59 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 2.0 m/s . In all parts, use the standard sign convention for direction. How long does it take the first stone to hit the water? What was the initial velocity of the second stone? What is the velocity of the first stone as it hits the water? What is the velocity of the second stone as it hits the water?

Explanation / Answer

1)
for 1st stone:
d = -59 m
vi = -2.0 m/s
a = -9.8 m/s^2

use:
d = vi*t + 0.5*a*t^2
-59 = -2.0*t - 0.5*9.8*t^2
-59 = -2.0*t - 4.9*t^2
4.9*t^2 + 2.0*t - 59 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 4.9
b = 2.0
c = -59

Roots can be found by
t = {-b + sqrt(b^2-4*a*c)}/2a
t = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 1.16*10^3

roots are :
t = 3.272 and t = -3.68

since t can't be negative, the possible value of t is
t = 3.272

Answer: 3.3 s

2)
for 2nd stone:
t = 3.3-1 = 2.3 s
d = -59 m
a = -9.8 m/s^2
use:
d = vi*t + 0.5*a*t^2
-59 = vi*2.3 - 0.5*9.8*(2.3)^2
-59 = vi*2.3 - 25.921
vi = -14.4 m/s
Answer: -14.4 m/s

3)
for 1st stone:
vf^2 = vi^2 + 2*a*d
vf^2 = 2.0^2 + 2*(-9.8)*(-59)
vf^2 = 4.0 + 1156.4
vf = -34.1 m/s (it is negative as it is directed downward)
Answer: -34.1 m/s

4)
for 2nd stone:
vf^2 = vi^2 + 2*a*d
vf^2 = 14.4^2 + 2*(-9.8)*(-59)
vf^2 = 207.36 + 1156.4
vf = -36.9 m/s (it is negative as it is directed downward)
Answer: -36.9 m/s

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