April releases a rock at rest from the top of a 40 m tower. Ignoring air resistance, what is the speed of the rock as it hits the ground? ( g = 9.8 m/s/s)

Calculate how long it take to hit the ground, by rearranging the equation

Y = (1/2) g t^2
which should be in your course/text materials.

Let Y be the 40 m distance that it falls. Solve for t.

on ce you have t, get the velocity from
V = gt

To determine the speed of the rock as it hits the ground, we can use the laws of motion and the concept of free fall.

First, let's denote the initial velocity of the rock as u (which is 0 m/s since it starts from rest). The final velocity of the rock when it hits the ground will be v, and the time it takes to reach the ground will be t.

Since the rock is subject to free fall, we can use the equation:

v = u + gt

Where:
- v is the final velocity of the rock.
- u is the initial velocity of the rock (0 m/s).
- g is the acceleration due to gravity (9.8 m/s²).
- t is the time it takes for the rock to reach the ground.

To find the time it takes for the rock to hit the ground, we can use the equation:

h = (1/2)gt²

Where:
- h is the height of the tower (40 m).
- g is the acceleration due to gravity (9.8 m/s²).
- t is the time it takes for the rock to reach the ground.

Rearranging this equation, we can solve for t:

t² = (2h)/g
t = √((2 × 40)/9.8)
t ≈ √(80/9.8)
t ≈ √8.16
t ≈ 2.86 seconds (rounded to two decimal places)

Now, we can substitute the value of t into the first equation to find the final velocity of the rock:

v = u + gt
v ≈ 0 + (9.8 × 2.86)
v ≈ 28 m/s

Therefore, the speed of the rock as it hits the ground is approximately 28 m/s.