height of cliff = h
height of top of second experiment = h + 2
Experiment 1:
h = (1/2) g t^2 = (1/2) g (4.15)^2
h = g (8.61 )
so g = h/8.61
Experiment 2
rise part:
initial speed up = Vi
v = Vi - g t
v = 0 in 2 meter rise
0 = Vi - g trise so trise = Vi/g
2 = Vi trise - (1/2) g (trise)^2
2 = Vi^2 /g - (1/2)Vi^2 /g
2 = (1/2) Vi^2/g
Vi^2 = 4 g
Vi = 2 sqrt g
y = height, 0 at the end
y = h + Vi t - (1/2)g t^2
0 = h + 2 sqrt g (6.3) - (1/2)g (6.3)^2
now from experiment 1
g = h/8.61
so we have
0 = h + 12.6 sqrt (h/8.61) -(1/2)(h/8.61)( 39.7)
0 = 4.3 h^.5 - 2.3 h
= h^.5 ( 4.3 - 2.3 h^.5)
h = 0 works of course but also
h^.5 = 1.87
h = 3.5 meters
An astronaut is stranded on another planet with UNKNOWN GRAVITY. He conducts two experiments. In the first, he drops a rock from a cliff and he finds that it takes the rock 4.15 seconds to reach the ground at the base of the cliff. In the second experiment, he throws the rock straight up to a heights he determines to be 2 meters before it falls to the ground at the base of the cliff with the first rock. The time for the second rock to reach the ground is 6.3 seconds. How high is the cliff?
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