Freefall


by motionman04
Tags: freefall
motionman04
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#1
Sep13-04, 05:44 PM
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I'm having some problems with this question, A ball is thrown directly downward, with an initial speed of 8.25 m/s, from a height of 29.4 m. After what time interval does the ball strike the ground?

I tried 29.4 + 8.25 m/s(x) + 1/2(-9.8m/s)(x), but that didn't turn out to be right. Can I get some help with this one?
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Pyrrhus
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Sep13-04, 05:49 PM
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Well in a xy coordinate system, initial position will be its height, and when it hits the ground it will have a position of 0, so it's final position must be 0
Doc Al
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#3
Sep13-04, 06:34 PM
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Quote Quote by motionman04
I tried 29.4 + 8.25 m/s(x) + 1/2(-9.8m/s)(x), but that didn't turn out to be right. Can I get some help with this one?
I assume you are trying to apply the following kinematic equation:
y = y_0 + v_0 t + (1/2)a t^2
Be sure to use a consistent sign convention: not only is the acceleration negative (a = - 9.8 m/s^2), don't forget that the initial velocity is also negative since it is thrown downward.
And, as Cyclovenom points out, the final postion is where y = 0. Solve for t.

Integral
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Sep13-04, 06:59 PM
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Freefall


29.4 + 8.25 m/s(x) + 1/2(-9.8m/s)(x),
Don't forget that the x (I would prefer t!) in the bold quantity needs to be squared.


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