A rock is thrown straight downward from a tree limb with an initial velocity v0. The rock has constant downward acceleration of 10 m/s2 during its fall. The rock moves 9 times as far during the first 3 seconds as during the first 1 second of fall. What is the value (magnitude only) of v0?
Let's denote the initial velocity of the rock as v0.
During the first second of fall, the rock travels a distance of d1 = (1/2)gt^2 = (1/2)(10 m/s^2)(1 s)^2 = 5 meters.
During the first three seconds of fall, the rock travels a distance of d2 = (1/2)g(3 s)^2 = 45 meters.
We know that the rock moves 9 times as far during the first 3 seconds as during the first 1 second of fall. Therefore:
d2 = 9d1
45 = 9d1
d1 = 5 meters
Now we can use the distance formula for falling objects to find v0:
d2 = (1/2)g(t2)^2 + v0t2
45 = (1/2)(10 m/s^2)(3 s)^2 + v0(3 s)
45 = 45 m + 3v0
v0 = (45 - 45 m)/3 s
v0 = 0 m/s
Therefore, the initial velocity of the rock is zero. This makes sense since the rock is thrown straight downward, and its velocity is entirely in the downward direction.
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