Wednesday

April 16, 2014

April 16, 2014

Posted by **Dan** on Tuesday, March 26, 2013 at 5:25pm.

vx = +225 m/s

at a height of

y = 920 m

above level ground as shown in the figure below when it releases a package. Ignoring air resistance, how much time will it take the package to reach the ground? (Express your answer to the nearest tenth of a second.)

- physics -
**Elena**, Tuesday, March 26, 2013 at 5:33pmy=gt²/2

t=sqrt(2y/g) =sqrt(2•920/9.8)=13.7 s

- physics -
**John**, Tuesday, March 26, 2013 at 5:34pmFor this question you're going to need a kinematic equation. Because you're solving for time, I would use d=Voy*t+0.5*a*t^2

The acceleration is 9.8 m/s^2 (making "down" positive) due to gravity.

There is no initial velocity (V) in the y direction.

You do not need to use the fact that Vox is 225 m/s.

So if d=Voy*t+0.5*a*t^2 first substitute the values that you know. You get 920=0.5*(9.8)*t^2

solve for t. You should get something like 13.7 seconds.

Hope that helps!

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