Question - Calculating the Maximum Height Reached by a Ball Thrown Vertically Upward

To find the maximum height \( h \) reached by the ball, we use the kinematics equation which relates the final velocity \( v \), initial velocity \( u \), acceleration \( a \), and displacement \( s \) (in this case, height \( h \)):

\[ v^2 = u^2 + 2as \]

At the maximum height, the final velocity \( v \) will be 0 m/s (since the ball stops rising before it starts to fall). The initial velocity \( u \) is 10 m/s (upward), and the acceleration \( a \) due to gravity is -10 m/s² (downward). Thus, we can plug these into our equation:

\[ 0 = (10)^2 + 2(-10)h \]

Which simplifies to:

\[ 0 = 100 - 20h \]

Now we solve for \( h \):

\[ 20h = 100 \]

\[ h = \frac{100}{20} \]

\[ h = 5 \]

Thus, the maximum height \( h \) reached by the ball is 5 meters.