To determine the height to which the ball reaches, we can use the principles of kinematics. Specifically, we'll use one of the kinematic equations related to the motion under uniform acceleration.
Let's outline the variables:
- Initial velocity ([tex]\( u \)[/tex]) = 10 m/s
- Final velocity at the highest point ([tex]\( v \)[/tex]) = 0 m/s (since the ball will momentarily stop before falling back down)
- Acceleration ([tex]\( a \)[/tex]) = -9.8 m/s² (the negative sign indicates that it is directed downward due to gravity)
We can use the following kinematic equation:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Rearrange the equation to solve for [tex]\( s \)[/tex] (displacement or height in this context):
[tex]\[ 0 = u^2 + 2as \][/tex]
[tex]\[ 0 = (10)^2 + 2(-9.8)s \][/tex]
[tex]\[ 0 = 100 - 19.6s \][/tex]
Now, isolate [tex]\( s \)[/tex]:
[tex]\[ 19.6s = 100 \][/tex]
[tex]\[ s = \frac{100}{19.6} \][/tex]
[tex]\[ s ≈ 5.1 \, \text{m} \][/tex]
Therefore, the height to which the ball goes is approximately [tex]\(\mathbf{5.1 \, \text{m}}\)[/tex].
Given the options:
A. 10 m
B. 5.1 m
C. 3.2 m
D. 7.5 m
The correct answer is B. 5.1 m.
