## Related questions with answers

(a) A stone is thrown upward with a certain speed on a planet where the free-fall acceleration is double that on Earth. How high does it rise compared to the height it rises on Earth? (b) If the initial speed were doubled, what change would that make?

Solution

Verified$\text{\textcolor{#c34632}{(a.)}}$ The maximum height can attain by the stone when thrown vertically upward on Earth's surface can be calculated as

$\begin{align*} v_y^2 = v_{0y}^2 -2g(y-y_0) \hspace{5mm} \implies (y-y_0) = \dfrac{v_{0y}^2 - v_y^2}{2g} \end{align*}$

Since the stone is thrown vertically upward, Hence $y_0 = 0$ and $v_y = 0$ at maximum height, Simplifying the equation gives

$(y-0) = \dfrac{v_{0y}^2 - 0^2}{2g} \hspace{5mm} \implies y = \dfrac{v_{0y}^2}{2g}$

We assume that on the other planet, the value of free fall acceleration is doubled compared that on Earth. So, the height it rises on this planet yields

$y_p = \dfrac{v_{0y}^2}{2(2\cdot g)} \hspace{5mm} \implies y_p = \dfrac{v_{0y}^2}{4g}$

The height on the other planet can compare to the height it rises on Earth by taking its ratio. Thus,

$\begin{align*} \dfrac{ y_p}{ y_E}& = \dfrac{\left(\dfrac{v_{0y}^2}{4g}\right) }{\left(\dfrac{v_{0y}^2}{2g}\right) } = \dfrac{\dfrac{1}{4}\cdot \left(\dfrac{v_{0y}^2}{g}\right) }{\dfrac{1}{2} \cdot \left(\dfrac{v_{0y}^2}{g}\right) }\\ & = \dfrac{1}{2} \end{align*}$

Thus, the height on other planet is one-half the value compared to the height of the stone rises on Earth.

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