Question

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?

Answer 1

Answer 2

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.