Notes: A ball is thrown upward with an initial velocity of 9.8 m/s. How high does it reach before it starts descending? Choose only one from the...

Monday, 18 May 2015

A ball is thrown upward with an initial velocity of 9.8 m/s. How high does it reach before it starts descending? Choose only one from the...

To solve, apply the third formula.

$\displaystyle{{v}_{{{f{{y}}}}}^{{2}}}={{v}_{{{i}{y}}}^{{2}}}+{2}{g{{d}}}_{{y}}$

Take note that when the ball reaches the maximum height, its velocity is zero. So plugging in the values

$\displaystyle{v}_{{{i}{y}}}={9.8}$ m/s

$\displaystyle{v}_{{{f{{y}}}}}={0}$

$\displaystyle{g{=}}-{9.8}$ m/s^2

the formula becomes

$\displaystyle{{0}}^{{2}}={{9.8}}^{{2}}+{2}{\left(-{9.8}\right)}{d}_{{y}}$

$\displaystyle{0}={96.04}-{19.6}{d}_{{y}}$

$\displaystyle{19.6}{d}_{{y}}={96.04}$

$\displaystyle{d}_{{y}}=\frac{{96.04}}{{19.6}}$

$\displaystyle{d}_{{y}}={4.9}$

Therefore, the maximum height of the ball is 4.9 meters.

To solve, apply the third formula.

$\displaystyle{{v}_{{{f{{y}}}}}^{{2}}}={{v}_{{{i}{y}}}^{{2}}}+{2}{g{{d}}}_{{y}}$

Take note that when the ball reaches the maximum height, its velocity is zero. So plugging in the values

$\displaystyle{v}_{{{i}{y}}}={9.8}$ m/s

$\displaystyle{v}_{{{f{{y}}}}}={0}$

$\displaystyle{g{=}}-{9.8}$ m/s^2

the formula becomes

$\displaystyle{{0}}^{{2}}={{9.8}}^{{2}}+{2}{\left(-{9.8}\right)}{d}_{{y}}$

$\displaystyle{0}={96.04}-{19.6}{d}_{{y}}$

$\displaystyle{19.6}{d}_{{y}}={96.04}$

$\displaystyle{d}_{{y}}=\frac{{96.04}}{{19.6}}$

$\displaystyle{d}_{{y}}={4.9}$

Therefore, the maximum height of the ball is 4.9 meters.

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