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# For a flourish at the end of her act, a juggler tosses a single ball high in the air. She catches the ball $3.2$ s later at the same height from which it was thrown. What was the initial upward speed of the ball?

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So basically we have two persons on the scene one of them is tossing the ball and the other is catching it later on at the same height after $t$= 3.2 seconds, using the symmetry of free fall to find the initial speed with which the ball was tossed, we use the following equation

$v_o=\dfrac{1}{2} ~ gt$

Where,

$g$: is acceleration due to gravity and it is 9.81 $\dfrac{\mathrm{m}}{\mathrm{s}^2}$

Keeping in mind that this equation can be only used provided that the act of tossing and catch was performed at the same height.

$v_o=\dfrac{1}{2} ~ (9.81 ~ \dfrac{\mathrm{m}}{\mathrm{s}^2} \times 3.2 \mathrm{~ sec} )= 15.696 ~ \dfrac{\mathrm{m}}{\mathrm{s}}$

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