Similar balls are thrown vertically each with a velocity `20ms^(-1)`, one on the surface of earth and the other on the surface of moon. What will be ratio of the maximum heights attained by them? (Acceleration on moon `= 1.7 ms^(-2)` approx)

Options

A. 6
B. `(1)/(6)`
C. `(1)/(5)`
D. 4

Answer

To solve the problem of finding the ratio of the maximum heights attained by two balls thrown vertically with the same initial velocity on the Earth and the Moon, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial velocity (u) = 20 m/s (for both balls)
- Acceleration due to gravity on Earth (g_e) = 10 m/s² (approximate value)
- Acceleration due to gravity on Moon (g_m) = 1.7 m/s²

2. **Use the Kinematic Equation:**
The maximum height (h) attained by an object thrown vertically can be calculated using the equation:
\[ v^2 = u^2 - 2gh \]
At maximum height, the final velocity (v) is 0. Therefore, the equation simplifies to:
\[ 0 = u^2 - 2gh \implies h = \frac{u^2}{2g} \]

3. **Calculate Maximum Height on Earth (h_e):**
Using the equation for maximum height:
\[ h_e = \frac{u^2}{2g_e} = \frac{(20)^2}{2 \times 10} = \frac{400}{20} = 20 \text{ m} \]

4. **Calculate Maximum Height on Moon (h_m):**
Similarly, for the Moon:
\[ h_m = \frac{u^2}{2g_m} = \frac{(20)^2}{2 \times 1.7} = \frac{400}{3.4} \approx 117.65 \text{ m} \]

5. **Find the Ratio of Maximum Heights (h_e : h_m):**
Now, we can find the ratio of the maximum heights attained:
\[ \text{Ratio} = \frac{h_e}{h_m} = \frac{20}{117.65} \approx 0.170 \]

6. **Express the Ratio in Simplified Form:**
To express the ratio in a simpler form, we can approximate:
\[ \frac{20}{117.65} \approx \frac{1}{6} \]

### Final Answer:
The ratio of the maximum heights attained by the balls on Earth and the Moon is approximately \( \frac{1}{6} \).

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