## Related questions with answers

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?

Solution

VerifiedSo 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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