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3 years ago

A stone is thrown straight upward with a speed of 20m/s. it is caught on its way down at a pint 5.00m above where it was thrown

Physics

1 answer:

Lorico [155]3 years ago

4 0

Answer:

<< how fast is it going when it was caught? >>

Working formula is

Vf^2 - Vo^2 = 2gH

where

Vf = velocity of stone when it was caught

Vo = initial velocity = 20 m/sec.

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

H = height at which stone was caught = 5 meters (given)

Substituting values,

Vf^2 - 20^2 = 2(9.8)(-5)

Vf^2 = 400 - 98

Vf^2 = 302

Vf = 17.38 m/sec.

<< how long did the trip take? >>

Formula is

H = Vo(T) + (1/2)gT^2

-5 = -20T + (1/2)(9.8)T^2

4.9T^2 - 20T + 5 = 0

Using the quadratic formula,

T = 3.81 seconds

Hope this helps.

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4 years ago

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I attached a diagram that shows how this force aligns with the force of gravity.
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4 years ago

A cube has a drag coefficient of 0.8. what would be the terminal velocity of a sugar cube 1 cm on a side in air (ρ = 1.2 kg/m3)?

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