To find out how high the ball goes when it is thrown straight up into the air with an initial speed of [tex]\(10 \, m/s\)[/tex], we can use one of the kinematic equations of motion. In this situation, we’ll use the equation concerning the maximum height when the final velocity is zero.
The kinematic equation for maximum height achieved is:
[tex]\[
v^2 = u^2 + 2as
\][/tex]
Where:
- [tex]\( v \)[/tex] is the final velocity (which is [tex]\( 0 \, m/s \)[/tex] at the highest point because the ball momentarily comes to rest before falling back down),
- [tex]\( u \)[/tex] is the initial velocity ( [tex]\(10 \, m/s\)[/tex] ),
- [tex]\( a \)[/tex] is the acceleration (in this case, due to gravity, which is negative because it is acting downward, so [tex]\( a = -9.8 \, m/s^2 \)[/tex]),
- [tex]\( s \)[/tex] is the displacement or the maximum height in this case (which we need to find).
Since the final velocity [tex]\( v \)[/tex] at the highest point is [tex]\( 0 \, m/s \)[/tex], we rearrange the equation to solve for [tex]\( s \)[/tex]:
[tex]\[
0 = u^2 + 2as
\][/tex]
Rewriting this equation for [tex]\( s \)[/tex]:
[tex]\[
s = \frac{-u^2}{2a}
\][/tex]
Substitute the known values [tex]\( u = 10 \, m/s \)[/tex] and [tex]\( a = -9.8 \, m/s^2 \)[/tex]:
[tex]\[
s = \frac{-(10 \, m/s)^2}{2 \times (-9.8 \, m/s^2)}
\][/tex]
Simplify the equation:
[tex]\[
s = \frac{-100}{-19.6}
\][/tex]
[tex]\[
s = \frac{100}{19.6}
\][/tex]
[tex]\[
s \approx 5.1020408163265305 \, m
\][/tex]
Therefore, the maximum height the ball reaches is approximately [tex]\( 5.1 \, m \)[/tex].
So, the correct answer is:
A. [tex]\( 5.1 \, m \)[/tex]
