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Step-by-Step Solution
Step 1: Identify the known variables
• Let the maximum height reached by the ball be $h$.
• At half the maximum height $\left(\frac{h}{2}\right)$, the speed of the ball is $10\,\text{m/s}$.
• At the maximum height $(h)$, the speed becomes $0\,\text{m/s}$.
• Acceleration due to gravity $(g)$ is $10\,\text{m/s}^2$ (downwards).
Step 2: Write down the relevant kinematic equation
For vertical motion under constant acceleration, we use the equation:
$v^2 = u^2 - 2 g \Delta y,
where $v$ is the final velocity, $u$ is the initial velocity, $g$ is the acceleration due to gravity, and $\Delta y$ is the change in height.
Step 3: Apply the equation at maximum height
At the maximum height $h$, the velocity $v = 0$. The initial velocity at $\frac{h}{2}$ is given as $10\,\text{m/s}$. Applying the equation from the point at $\frac{h}{2}$ to the top:
$0 = (10)^2 - 2 \times 10 \times \left(h - \frac{h}{2}\right).$
Here, $\left(h - \frac{h}{2}\right)$ represents the displacement from the point at half height to the top of the trajectory.
Step 4: Simplify and solve for h
Simplifying,
$0 = 100 - 2 \times 10 \times \frac{h}{2} = 100 - 10h.
Thus,
$10h = 100 \quad \Rightarrow \quad h = 10\,\text{m}.
Step 5: Conclude the maximum height
Therefore, the ball rises to a maximum height of $10\,\text{m}$.
