To determine how fast the brick is going when it hits the ground, we can use the principles of physics. Specifically, we will use the equation of motion that relates the final velocity, initial velocity, acceleration, and distance covered:
[tex]\[ v^2 = u^2 + 2gh \][/tex]
Where:
- [tex]\( v \)[/tex] is the final velocity (which we want to find).
- [tex]\( u \)[/tex] is the initial velocity. Since the brick is released from rest, [tex]\( u = 0 \)[/tex].
- [tex]\( g \)[/tex] is the acceleration due to gravity, which is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
- [tex]\( h \)[/tex] is the height from which the brick is released, which is [tex]\( 3 \, \text{meters} \)[/tex].
1. First, we start with the given equation and plug in the initial velocity [tex]\( u = 0 \)[/tex]:
[tex]\[ v^2 = 0 + 2gh \][/tex]
2. Substitute [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex] and [tex]\( h = 3 \, \text{meters} \)[/tex]:
[tex]\[ v^2 = 2 \times 9.8 \times 3 \][/tex]
3. Perform the multiplication inside the square root:
[tex]\[ v^2 = 2 \times 9.8 \times 3 \][/tex]
[tex]\[ v^2 = 58.8 \][/tex]
4. Now, take the square root of both sides to solve for [tex]\( v \)[/tex]:
[tex]\[ v = \sqrt{58.8} \][/tex]
5. The square root of 58.8 is approximately:
[tex]\[ v \approx 7.668 \][/tex]
So, the speed of the brick when it hits the ground is approximately [tex]\( 7.668 \, \text{m/s} \)[/tex].
Comparing this calculated value with the given options:
A. [tex]\( 8.1 \, \text{m/s} \)[/tex]
B. [tex]\( 7.2 \, \text{m/s} \)[/tex]
C. [tex]\( 7.7 \, \text{m/s} \)[/tex]
D. [tex]\( 9.9 \, \text{m/s} \)[/tex]
The closest and most accurate option is:
C. [tex]\( 7.7 \, \text{m/s} \)[/tex]
