Certainly! Let's solve this problem step-by-step.
1. Given Data:
- Height of the tower (h) = 40 meters
- Acceleration due to gravity (g) = 9.8 m/s²
- Initial velocity (u) of the rock = 0 m/s (since it is released from rest)
2. What We Need to Find:
- The final speed (v) of the rock as it hits the ground
3. Approach:
We can use one of the kinematic equations of motion that relate initial velocity, final velocity, acceleration, and displacement. The equation we’ll use is:
[tex]\[
v^2 = u^2 + 2gh
\][/tex]
- [tex]\( v \)[/tex] is the final velocity we're trying to find
- [tex]\( u \)[/tex] is the initial velocity, which is 0
- [tex]\( g \)[/tex] is the acceleration due to gravity, 9.8 m/s²
- [tex]\( h \)[/tex] is the height, 40 meters
4. Substitute the Given Values:
Since [tex]\( u = 0 \)[/tex]:
[tex]\[
v^2 = 0^2 + 2 \cdot 9.8 \cdot 40
\][/tex]
Simplifying inside the parentheses first:
[tex]\[
v^2 = 2 \cdot 9.8 \cdot 40
\[
\[
v^2 = 784
\][/tex]
5. Solve for [tex]\( v \)[/tex]:
To find [tex]\( v \)[/tex], we take the square root of both sides:
[tex]\[
v = \sqrt{784}
\[
\
6. Final Answer:
\[
v = 28 \, \text{m/s}
\][/tex]
So, the speed of the rock as it hits the ground is 28 m/s. Therefore, the correct answer is:
- 28 m/s
