A rock is tossed straight up with a speed of 29m/s. When it returns, it falls into a hole 13m deep.

Solve for rock's velocity as it hits the bottom of the hole.

&How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

(1/2)M V2^2 = (1/2)MV1^2 + M g D

D is the depth of the hole.
V1 is the initial "toss" velocity
V2 is the velocity at the bottom of the hole.
Cancel out the M's and solve for V2.

For time in the air, solve the following equation for t:

Y = -13 = V1*t - (g/2)*t^2
g = 9.8 m/s^2

To solve for the rock's velocity as it hits the bottom of the hole, we need to consider the motion of the rock as it goes up and comes back down. Let's break down the problem into two parts: the upward motion and the downward motion.

1. Upward motion:
We know that the initial velocity is 29 m/s since the rock is tossed straight up. The force of gravity acts against the motion of the rock, which eventually brings it to a stop at the highest point. At this highest point, the rock's velocity becomes zero.
Using the formula for velocity in freefall motion, we can find the time it takes for the rock to reach the highest point using the equation:
v = u + at
0 = 29 - 9.8t (where acceleration due to gravity, a = -9.8 m/s^2)
Solving for t, we get t = 2.96 seconds.

2. Downward motion:
When the rock reaches its highest point, it starts falling down due to the force of gravity. The initial velocity at this point is zero, and the distance it needs to travel is the sum of the initial height and the depth of the hole, which is 13m.
Using the equation for distance traveled in freefall motion, we can find the time it takes for the rock to fall into the hole:
s = ut + (1/2)at^2
13 = 0 + (1/2)(-9.8)t^2
26 = -4.9t^2
t^2 = 26 / 4.9
t^2 = 5.31
t ≈ 2.3 seconds

So the rock is in the air for a total of approximately 2.96 seconds (for the upward motion) + 2.3 seconds (for the downward motion) ≈ 5.26 seconds.

To calculate the rock's velocity as it hits the bottom of the hole, we can use the equation for velocity in freefall motion:
v = u + at
v = 0 + (-9.8)(2.3) (as gravity acts downward)
v ≈ -22.54 m/s (since the negative sign indicates the downward direction)

Therefore, the rock's velocity as it hits the bottom of the hole is approximately 22.54 m/s downward.