Gerrard want to know the depth of a well. Since he remebers his Physics, he prop
ID: 1358852 • Letter: G
Question
Gerrard want to know the depth of a well. Since he remebers his Physics, he proposes to calculate the depth of the well by measuring the time it takes for a rock to fall to the bottom. He drops the rock and 5.7 seconds later hears the splash. How deep is the well? (answer in "m") if the well were 1437 m deep, at what speed will the rock hit the bottom of the well? (answer in "m/s") I SUBMITTED THIS PROBLEM OVER 30 MIN AGO AND RECEIVED INCORRECT ANSWERS OF : 137.6M for 1st part and 167.8 for second part of question..
Explanation / Answer
a)
here by using the second equation of motion
y = u * t + 0.5 * a * t^2
y = 0 + 0.5 * 9.8 * 5.7^2
y = 159.2 m
the depth of the well is 159.2 m
b)
if the well is 1437 m deep then
by using the third equation of motino
v^2 - u^2 = 2 * a *s
v^2 - 0 = 2 * 9.8 * 1437
v = sqrt( 2 * 9.8 * 1437 )
v = 167.8 m/s
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