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Step-by-Step Solution
Step 1: Identify the Key Information
• The ball is thrown vertically down from a height of 20 m with an initial velocity $v_0$.
• It collides with the ground and loses 50% of its energy in the collision.
• Despite losing 50% of its energy, the ball rebounds back to the same height (20 m).
• Gravitational acceleration $g = 10$ m/s2.
Step 2: Relate Energy Before and After Collision
• Initial total energy (at the top, just before it starts moving down) is:
$$
E_\text{initial} = mg \times 20 + \frac{1}{2} m {v_0}^2.
$$
Here, $mg \times 20$ is the potential energy at height 20 m and $\frac{1}{2} m {v_0}^2$ is the initial kinetic energy due to velocity $v_0$.
• Just before hitting the ground, all the initial potential and kinetic energy has converted into kinetic energy:
$$
E_\text{just before ground} = \frac{1}{2} m v^2.
$$
(Here, $v$ is the speed just before impact.)
Step 3: Analyze the Effect of Losing 50% of Energy
• The ball loses 50% (half) of its energy on collision. Thus, immediately after collision, the energy becomes:
$$
E_\text{after collision} = \frac{1}{2} \times E_\text{just before ground}.
$$
• After collision, the ball still rises back to the same height (20 m), so its energy during the ascent must equal:
$$
E_\text{after collision} = mg \times 20.
$$
• Therefore,
$$
\frac{1}{2} \times \left(\frac{1}{2} m v^2\right) = mg \times 20
\quad\Longrightarrow\quad
\frac{1}{4} m v^2 = mg \times 20.
$$
Simplifying,
$$
v^2 = 80g.
$$
Step 4: Determine $v_0$ Using Energy Conservation
• From the top to just before ground:
$$
mg \times 20 + \frac{1}{2} m {v_0}^2 = \frac{1}{2} m v^2.
$$
• Substitute $v^2 = 80g$ into this:
$$
mg \times 20 + \frac{1}{2} m {v_0}^2 = \frac{1}{2} m \times 80g.
$$
Simplify:
$$
20mg + \frac{1}{2} m {v_0}^2 = 40mg.
$$
$$
\frac{1}{2} m {v_0}^2 = 20mg
\quad\Longrightarrow\quad
{v_0}^2 = 40g.
$$
• Putting $g = 10$ m/s2:
$$
{v_0}^2 = 40 \times 10 = 400,
\quad\Longrightarrow\quad
v_0 = 20 \,\text{m/s}.
$$
Step 5: Conclusion
Therefore, the initial velocity $v_0$ with which the ball is thrown vertically downward is
20 m/s.
