Question

Chef Susan tosses some pizza dough in the air, then catches it again at the same height. The pizza dough is in the air for 1.0\,\text s1.0s1, point, 0, start text, s, end text. We can ignore air resistance. What was the dough's velocity at the moment it was tossed into the air?

Answer 1

Answer:

The velocity of the release is 4.9 m/s.

Explanation:

Susan catches the dough, at the point of the release only therefore, the displacement of the piece of dough is 0 and the total time elapsed is 1s.

The gravity exerted will be only due to gravity as air resistance is neglected, so lets' assign downwards direction as negative and the gravity will come out to be -g.

We have to apply Second Equation of Motion which states that,

$s=ut+\frac{1}{2} at^{2}$

s=0

t=1

a=-g=-9.8

u=??

Plugging in the values, we get

0=u-4.9

u=4.9 m/s

Therefore, the velocity of the release is 4.9 m/s