Complicated Physic question.?
An alert hiker sees a boulder fall from the top of a distant cliff and notes that it takes 1.30 s for the boulder to fall the last third of the way to the ground. You may ignore air resistance.
This is three times the distance that the boulder would fall in 1.3 seconds if dropped from rest (as if it fell for 1.3 seconds, came to a stop, fell again, stopped, etc). However, the boulder keeps picking up speed as it falls, so the distance it falls is much greater than this value.
I've search for the same question that other user posted in Yahoo Answer, but none of the question are correct.
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Verified answer
t - time to complete the fall
x - total fall distance
x = 0.5 * g * (t^2)
we also know
0.667*x = 0.5 * g * [t-1.3]^2
Now you have two equations you can solve
0.667 * [0.5*g*t^2] = 0.5*g*[t-1.3]^2
0.667 * t^2 = t^2 - 2.6t + 1.69
0.333t^2 - 2.6t + 1.69 = 0
t = 2.6 or 5.2 secs
x = 33.12 or 132.49 m