A rock is tossed straight up with a velocity of 33.9 m/s. When it returns, it falls into a hole 11.9 m deep. What is the rock's velocity as it hits the bottom of the hole?

Check my previous answer to this same question, earlier today.

http://www.jiskha.com/display.cgi?id=1286206039

To find the rock's velocity as it hits the bottom of the hole, we need to consider the two separate phases of its motion - the upward trajectory and the downward trajectory.

First, let's find out how high the rock goes before falling back down. We can use the kinematic equation:

vf^2 = vi^2 + 2ad

where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the displacement.

Since the rock is tossed straight up, its final velocity at the highest point will be zero (vf = 0). The initial velocity is given as 33.9 m/s, and the displacement is the maximum height the rock reaches, which is the same as the depth of the hole, -11.9 m (negative because it is below the starting point).

Plugging these values into the equation, we have:

0^2 = (33.9 m/s)^2 + 2(-9.8 m/s^2)(-11.9 m)

Simplifying:

0 = 1148.21 + 231.144

Rearranging the equation, we find the maximum height (distance above the starting point) the rock reaches is:

d = (vi^2) / (2a)

d = (33.9 m/s)^2 / (2 * (-9.8 m/s^2))

d ≈ 61.736 m

Now, let's find the final velocity when the rock hits the bottom of the hole. We can use the same kinematic equation, but this time with a positive displacement value (11.9 m) since it is below the starting point:

vf^2 = vi^2 + 2ad

Here, the final velocity (vf) is what we're trying to find, the initial velocity (vi) is zero (since the rock starts from rest at the highest point), the acceleration (a) is -9.8 m/s^2 (gravity pulling the rock downwards), and the displacement (d) is 11.9 m.

Plugging in these values:

vf^2 = 0 + 2(-9.8 m/s^2)(11.9 m)

Simplifying:

vf^2 = -233.144

Taking the square root:

vf ≈ -15.26 m/s (negative sign indicates downward direction)

So, the rock's velocity as it hits the bottom of the hole is approximately 15.26 m/s downward.