An object is thrown vertically upward with a certain initial velocity in a world where the acceleration due to gravity is `19.6 m//s^(2)`. The height to which it rises is _____ that to which the object would rise if thrown upward with the same initial velocity on the Earth. Neglect air resistance.

Options

A. `1.77 m//s^(2)`
B. `3.60 m//s^(2)`
C. `2.98 m//s^(2)`
D. `7.36 m//s^(2)`

Answer

To solve the problem, we need to compare the heights reached by an object thrown vertically upward with the same initial velocity in two different gravitational fields: one with an acceleration due to gravity of `19.6 m/s²` and the other with `9.8 m/s²` (Earth's gravity).

### Step 1: Understand the relationship between height and acceleration due to gravity
When an object is thrown upward, it will rise to a maximum height where its final velocity becomes zero. The height (H) attained by the object can be determined using the equation of motion:

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

Where:
- \( v \) = final velocity (0 m/s at maximum height)
- \( u \) = initial velocity
- \( a \) = acceleration (negative for gravity)
- \( H \) = height attained

### Step 2: Set up the equations for both scenarios
1. For the world with \( g = 19.6 m/s² \):
\[ 0 = u^2 - 2(19.6)H_1 \]
Rearranging gives:
\[ H_1 = \frac{u^2}{2 \times 19.6} \]

2. For Earth with \( g = 9.8 m/s² \):
\[ 0 = u^2 - 2(9.8)H_2 \]
Rearranging gives:
\[ H_2 = \frac{u^2}{2 \times 9.8} \]

### Step 3: Compare the heights \( H_1 \) and \( H_2 \)
From the equations derived:
- \( H_1 = \frac{u^2}{39.2} \)
- \( H_2 = \frac{u^2}{19.6} \)

### Step 4: Establish the relationship between \( H_1 \) and \( H_2 \)
To find the relationship between \( H_1 \) and \( H_2 \):
\[ \frac{H_1}{H_2} = \frac{\frac{u^2}{39.2}}{\frac{u^2}{19.6}} = \frac{19.6}{39.2} = \frac{1}{2} \]

Thus, we find that:
\[ H_1 = \frac{1}{2} H_2 \]

### Conclusion
The height to which the object rises in the world with \( g = 19.6 m/s² \) is half that of the height it would rise if thrown upward with the same initial velocity on Earth.

### Final Answer
The height to which it rises is **half** that to which the object would rise if thrown upward with the same initial velocity on the Earth.

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