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?

Well, let's hope this rock doesn't have any fear of heights or else it might chicken out on its way down!

To solve this problem, we need to find the final velocity of the rock as it hits the bottom of the hole. We know that the initial velocity when the rock was thrown straight up was 33.9 m/s. But as the rock reaches its highest point, its velocity becomes zero since it momentarily stops before coming back down.

Now, when the rock falls back down into the hole, it will accelerate due to gravity. We also know that the depth of the hole is 11.9 m.

So, using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, we can plug in the values.

We already know that u = 33.9 m/s, a = 9.8 m/s^2 (acceleration due to gravity), and s = 11.9 m.

Solving for v, we have v^2 = 33.9^2 + 2(-9.8)(11.9).

After some calculations, we find that v^2 ≈ 383.66.

Taking the square root of both sides, we have v ≈ 19.58 m/s.

So, the velocity of the rock as it hits the bottom of the hole is approximately 19.58 m/s.

I guess we can say that gravity really pushed the rock to reach great speeds...but it's a good thing it didn't land on our heads!

To find the rock's velocity as it hits the bottom of the hole, we need to consider the motion of the rock while it's in the air.

First, let's determine the time it takes for the rock to reach its maximum height. We can use the fact that the vertical motion of a projectile can be described by the equation:

y = y₀ + v₀y⋅t - 1/2⋅g⋅t²

Where:
- y is the vertical position (which in this case is the height)
- y₀ is the initial vertical position (which is 0 because we're measuring from the ground)
- v₀y is the initial vertical velocity (33.9 m/s in the upward direction)
- g is the acceleration due to gravity (9.8 m/s²)
- t is the time

Since the rock is at maximum height when it reaches its peak, its final vertical velocity (v_fy) is zero. Therefore, we can rewrite the equation as:

0 = 0 + 33.9⋅sin(θ)⋅t - 1/2⋅9.8⋅t²

where θ is the angle of projection. Since the rock is thrown straight up, θ is 90°.

0 = 33.9⋅sin(90°)⋅t - 1/2⋅9.8⋅t²

0 = 0 + 33.9⋅t - 4.9⋅t²

Rearranging the equation, we get a quadratic equation:

4.9⋅t² - 33.9⋅t = 0

We can solve this equation to find the time it takes for the rock to reach its peak. By factoring out t, we have:

t(4.9t - 33.9) = 0

Setting each factor equal to zero and solving for t, we get two solutions:

t₁ = 0 (which is the initial time, t = 0)
t₂ = 33.9 / 4.9

Since t cannot be zero (as it represents the time when the rock reaches the peak), we take the positive value for t:

t = 33.9 / 4.9 ≈ 6.918 seconds

Now that we know it takes approximately 6.918 seconds for the rock to reach its maximum height, we can find the time it takes for the rock to fall back down. Since the total time of flight is double the time it takes to reach the peak, the time for the downward journey is:

t_down = 2⋅t ≈ 2⋅6.918 ≈ 13.836 seconds

During the downward journey, the rock is accelerated by gravity. The final velocity (v_f) can be calculated using the equation:

v_f = v₀ + g⋅t

where:
- v₀ is the initial velocity (which is the same as the final velocity during the upward journey, 33.9 m/s)
- g is the acceleration due to gravity (9.8 m/s²)
- t is the time for the downward journey (13.836 seconds)

Plugging in the values, we have:

v_f = 33.9 + 9.8⋅13.836 ≈ 33.9 + 135.6528 ≈ 169.5528

The rock's velocity as it hits the bottom of the hole is approximately 169.5528 m/s.