Question

5
1 point
A person working at a museum is placing a priceless vase on a new stand. The stand is 1.5m high and a speck of dust causes them to sneeze. Their sneeze tips the
vase over and it begins to fall. How long is the vase in the air before it crashes?

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Answer 1

Given:

A person working at a museum is placing a priceless vase on a new stand. The stand is 1.5m high and a speck of dust causes them to sneeze. Their sneeze tips the vase over and it begins to fall.

To find:

Time for which the vase is in air before it crashes.

Calculation:

Considering the vase to be a point object and assuming the gravitational acceleration to be constant :

Applying equations of kinematics:

$\therefore \: h = ut + \dfrac{1}{2} a{t}^{2}$

$= > \: h = ut + \dfrac{1}{2} g{t}^{2}$

$= > \: 1.5 = (0)t + \dfrac{1}{2} \times (10) \times {t}^{2}$

$= > \: 1.5 = \dfrac{1}{2} \times (10) \times {t}^{2}$

$= > \: 3 = 10 \times {t}^{2}$

$= > \: {t}^{2} = \dfrac{3}{10}$

$= > \: {t}^{2} = 0.3$

$= > \: t = 0.547 \: sec$

So, final answer is:

$\boxed{ \bf{ \large{\: time\:in\:air = 0.547 \: sec}}}$